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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 251. |
Given positive integers `r>1,n> 2, n` being even and the coefficient of `(3r)th` term and `(r+ 2)th` term in the expansion of `(1 +x)^(2n)` are equal; find rA. `n = 2r`B. `n = 3r`C. `n = 2r + 1`D. None of these |
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Answer» Given that, `r gt 1, n gt 2` and the coefficients of `(3r) th " and " (r + 2)th` term are equal in the expansion of `(1 + x)^(2n)` Then, `T_(3r) = T_(3r - 1 + 1) = .^(2n)C_(3r - 1) x^(3r - 1)` and `T_(r + 2) = T_(r + 1 + 1) = .^(2n)C_(r + 1) x^(r + 1)` Given, `.^(2n)C_(3r - 1) = .^(2n)C_(r + 1) " " [:. .^(n)C_(x) = .^(n)C_(y) rArr x + y = n]` `rArr 3 r - 1 + r + 1 = 2n` `rArr 4 r = 2n rArr n = (4r)/(2)` `.: n = 2r` |
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| 252. |
Let m be the smallest positive integer such that the coefficient of `x^2` in the expansion of `(1+x)^2 + (1 +x)^3 + (1 + x)^4 +........+ (1+x)^49 + (1 + mx)^50` is `(3n + 1) .^51C_3` for some positive integer n. Then the value of n is |
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Answer» Coefficient of `x^(2)` in expansion `= 1+.^(3)C_(2)+.^(4)C_(2)+.^(5)C_(2) + "….." + .^(49)C_(2)+.^(50)C_(2).m^(2)` [as `.^(n)C_(r)+.^(n)C_(r-1) = .^(n+1)C_(r)`] `= (.^(3)C_(5)+.^(3)C_(2)) + .^(4)C_(2) + .^(5)C_(2) + "…." + .^(49)C_(2) + .^(50)C_(2)m^(2)` `= (.^(4)C_(3) + .^(4)C_(2)) + "....." + .^(50)C_(2)m^(2)` `= .^(5)C_(3) + .^(50)C_(2)m^(2) + .^(50)C_(2)m^(2)` `= .^(50)C_(3) + .^(50)C_(2)m^(2)+.^(50)C_(2)-.^(50)C_(2)` `= .^(51)C_(3)+.^(50)C_(2)(m^(2)-1)` `= (3n+1).^(51)C_(3)` (given) `:. 3n.(51)/(3).^(50)C_(2) = .^(50)C_(2)(m^(2) - 1)` `(m^(2)-1)/(51) = n` Value of n is 5. |
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| 253. |
The last term in the binomial expansion of `(2^(1/3) -1/sqrt(2))^n` is `(1/(3.9^(1/3)))^(log_3 8)` then the 5th term from the beginning isA. `(1)/(2)xx""^(10)C_(6)`B. `2xx""^(10)C_(4)`C. `(1)/(2)xx""(10)C_(4)`D. none of these |
| Answer» Correct Answer - A | |
| 254. |
If [x] denotes the greatest integer less then or equal to x, then `[ (6sqrt(6) + 14)^(2n +1)]`A. is an even integerB. is an odd integerC. depands on nD. none of these |
| Answer» Correct Answer - a | |
| 255. |
Find the value of `{3^(2003)//28}, w h e r e{dot}`denotes the fractional part. |
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Answer» Correct Answer - `19//28` `E = 3^(2003) = 3^(2001) xx 3^(2) = 9(27)^(667) = 9(28-1)^(667)` `rArr E = 9[.^(667)C_(0) 28^(667) - .^(667)C_(1)(28)^(666) + "….."-.^(667)C_(667)]` `= 9 xx 28k - 9` `rArr E/28 = 9k - (9)/(28) = 9k = 1 + 19/28` That means if we divide `3^(2003)` by `28`, the remainder is `19`. Thus, `{(3^(2003))/(28)} = 19/28` |
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| 256. |
If in the expansion f `(1+x)^m(1-x)^n, the coefficient of x and `x^2` are 3 and -6 respectively then (A) m=9 (B) n=12 (C) m=12 (D) n=9A. 6B. 9C. 12D. 24 |
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Answer» `(1+x)^m(1-x)^n=[1+mx+(m(m-1))/(2)x^2+......]` `[1-nx+(n(n-1))/(2)x^2+.......]` `=1+(m-n)x+[(m(m-1))/(2)+(n(n-1))/(2)-mn]x^2+......` term containg power of `x ge 3`. Now `m-n=3 .....(i)` and `(1)/(2)m(m-1)+(1)/(2)n(n-1)-mn=-6`. `rArr m(m-1)+nn(n-1)-2mn=-12`. `rArr m^2-m+n^2-n-2mn=-12` `rArr (m-n)^2-(m+n)=-12` `rArr m=n=9+12=21....(ii)`. On solving Eqs.(i) and (ii), we get `m=12`. |
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| 257. |
If `n`is an even positive integer, then find the value of `x`if the greatest term in the expansion of `1+x`^n may have the greatest coefficient also. |
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Answer» Correct Answer - `x gt (n)/(n+2)` and `x lt (n+2)/(n)` If n is even, the greatest coefficient is `.^(n)c_(n//2)`. Therefore, the greatest term is `.^(n)C_(n//2)x^(n//2)`. `:. .^(n)C_(n//2)x^(n//2)gt.^(n)C_((n//2)-1)x^((n-2)//2)` and `.^(n)C_(n//2)x^(n//2) gt .^(n)C_((n//2)+1)x^((n//2)+1)` `rArr (n-n/2+1)/((n)/(2)) xgt 1` and `((n)/(2))/((n)/(2) + 1) x lt 1` `rArr xgt ((n)/(2))/(n/2+1)` and `x lt ((n)/(2)+1)/((n)/(2))` `rArr x ge (n)/(n+2)` and `x lt (n+2)/(n)` |
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| 258. |
If `p=(8+3sqrt(7))^n a n df=p-[p],w h e r e[dot]`denotes the greatest integer function, then the value of `p(1-f)`is equal to`1`b. `2`c. `2^n`d. `2^(2n)`A. 1B. 2C. `2^(n)`D. `2^(2n)` |
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Answer» Correct Answer - A `p=(8+3sqrt(7))^(n)=.^(n)C_(0)8^(n)+.^(n)C_(1)8^(n-1)(3sqrt(7))+"...."` Let `p_(1)=(8-3sqrt(7))^(n)=.^(n)C_(0)8^(n)-.^(7)C_(1)8^(n-1)(3sqrt(7))+"....."` `p + p_(1) = 2(.^(n)C_(0)8^(n)+.^(n)C_(2)8^(n-2)(3sqrt(7))^(2)+".....")="even integer"` `p_(1)` clearly belongs to `(0,1)` `rArr [p] = f+p_(1) = "even integer"` `rArr f + p_(1) = "integer"` `f in (0,1),p_(1)in(0,1)` `rArr f + p in (0,2)` `rArr f+ p_(1) = 1` `rArr p_(1)=1-f` Now, `p(1-f)=pp_(1)=[(8+3sqrt(7))^(n)(8-3sqrt(7))]^(n)=1` |
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| 259. |
In the binomial expansion of `(a-b)^n , ngeq5,`the sum of 5thand 6th terms is zero, then `a/b`equals(1) `5/(n-4)`(2) `6/(n-5)`(3) `(n-5)/6`(4) `(n-4)/5`A. `(n - 5)/(6)`B. `(n - 4)/(5)`C. `(5)/(n - 4)`D. `(6)/(n - 5)` |
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Answer» Given `T_5+T_6=0` `rArr .^Nc_4a^(n-4)b^4-.^nC_5a^(n-5)b^5=0` `rArr .^nC_4 a^(n-4)b^4=.^Nc_5 a^(n-5)b^5rArr (a)/(b)=(.^Nc_5)/(.^C_4)=(n-4)/(5)` |
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| 260. |
Find the 6th term of the expansion `(y^(1//2) + x^(1//3))^(n)` , if the binomial coefficient of 3rd term from the end is 45. |
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Answer» Correct Answer - `252y^(5//2)x^(5//3)` Coefficient of 3rd term from the beginning is always equal to the coefficient of 3rd term frm the end. Now, `T_(r+1) - .^(n) C(y^(1//2))^(n-r)*(x^(1//3))r.` Coefficient of `T_(3) = .^(n)C_(2) = (n(n-1))/2.` ` (n(n-1))/2 = 45 rArr n^(2) - n-90 = 0 rArr n^(2) - 10n + 9n - 90 = 0` `rArr ( n-10 (n+9) = 0 rArr n = 10` `:. T_(6) = T_(5+1) = .^(10)C_(5) (y^(1//2))^((10-5))*(x^(1//3))^(5)` `= (10 xx 9 xx 8 xx 7 xx 6)/(5 xx 4 xx 3 xx 2 xx 1) xxy^(5//2)x^(5//3) = 252y^(5//2)x^(5//3).` |
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| 261. |
Find the middle term in the expansion of `(x^2+1/(x^2)+2^n)dot` |
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Answer» Correct Answer - `((2n)!)/((n!)^(2))` `(x^(2)+1/(x^(2))+2)^(n) = (x+1/x)^(2n)` Hence, the middle term is `T_((2n)/(2)+1) = T_(n+1) = .^(2n)C_(n)(x)^(n)(1/x)^(n) = .^(2n)C_(n)= ((2n)!)/((n!)^(2))` |
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| 262. |
If `x=1//3,`find the greatest tem in the expansion of `(1+4x)^8dot` |
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Answer» Correct Answer - `T_(6) = 56(4/3)^(5)` We have Now, `(T_(r+1))/(T_(r)) = (n-r+1)/(r ) (4x) = (9-r)/(r ).(4)/(3)` (Putting `n = 8, x=1//3`) `:. (T_(r+1))/(T_(r)) ge 1` `rArr 36-4r ge 3r` `rArr 36 ge 7r` `rArr r le 51/7` Hence `r = 5`, and therefore, `T_(5+1)` i.e., `6^(th)` term is greatest. `rArr T_(6) = .^(8)C_(5)(4x)^(5) = (8!)/(5! xx 3!) .(4/3)^(5)` `= (8xx7xx6)/(6) (4/3)^(5)` `= 56(4/3)^(5)` |
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| 263. |
Let `aa n db`be the coefficients of `x^3`in `(1+x+2x^2+3x^2)^4a n d(1+x+2x^2+3x^3+4x^4)^4,`then respectively. Then the value of `4a//b`is. |
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Answer» Correct Answer - 4 We have `b=` coefficient of `x^(3)`in `((1+x+2x^(3)+3x^(3))+4x^(4))^(4)` = coefficient of `x^(3)` in `[.^(4)C_(0)(1+x+2x^(2)+3x^(3))^(4)(4x^(4))^(0) + .^(4)C_(1)(1+x+2x^(2)+3x^(3))^(3)(4x^(4))^(1)+"…"]` = coefficient of `x^(3)` in `(1+x+2x^(2)+3x^(3))^(4) = a` Hence `4a//b = 4`. |
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| 264. |
Find the greatest term in the expansion of `sqrt(3)(1+1/(sqrt(3)))^(20)dot`A. `(25840)/(9)`B. `(24840)/(9)`C. `(26840)/(9)`D. none of these |
| Answer» Correct Answer - a | |
| 265. |
If the 3rd, 4th , 5th and 6th term in the expansion of`(x+alpha)^n`be, respectively, `a ,b ,ca n dd ,`prove that `(b^2-a c)/(c^2-b d)=(5a)/(3c)dot` |
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Answer» We know that `(T_(r+1))/(T_(r)), (n-r+1)/(r)(alpha)/(x)` `T_(3)=a,T_(4)=b,T_(5)=c,T_(6)=d` Putting `r = 3, 4,5` in the abvoe we get `(T_(4))/(T_(3)) = (n-2)/(3)(alpha)/(x) = b/a` `(T_(5))/(T_(4)) = (n-3)/(4)(alpha)/(x) = c/b` `(T_(6))/(T_(5)) = (n-4)/(5) (alpha)/(x) = d/c` We have to prove `(b^(2)-ac)/(c^(2)-bd) = (5a)/(3c)` or `((b)/(c)-(a)/(b))/((c)/(b)-(d)/(c)) = (5a)/(3c)` Now, `((b)/(c) - (a)/(b))/((c)/(b)-(d)/(c)) = ((4x)/((n-3)alpha)-(3x)/((n-2)alpha))/(((n-3)alpha)/(4x) - ((n-4)alpha)/(5x))` `= (x^(2))/(alpha^(2))(4xx5)/((n-2)(n-3))(4n-8-3n+9)/(5n-15-4n+16)` `= (x^(2))/(alpha^(2))(4xx5)/((n-2)(n-3))` Also, `(5a)/(3c) = (x^(2))/(alpha^(2))(4xx5)/((n-2)(n-3))` `:. L.H.S. = R.H.S.` |
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| 266. |
Write the middle term in the expansion of `(x+1/x)^(10)dot`A. `""^(n)C_(1) (1)/(x)`B. `""^(10)C_(5)`C. `""^(10)C_(6)`D. `""^(10)C_(7) x` |
| Answer» Correct Answer - b | |
| 267. |
The largest value of `x`for which the fourth tem in the expansion `(5^2/3(log)_5sqrt(4^(x+44))+1/(5^(log)_5 2^((x-1)+7 3)))`is 336 is. |
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Answer» Correct Answer - 4 `(5^(2/5sqrt(4^(2)+44))+(1)/(5^(log2sqrt(2^(2-1))+7)))^(8)` `= ((sqrt(4^(4)+44)^(2//5))+((1)/(3sqrt(2^(x-1)+7))))^(8)` `= ((4^(x)+44)^(1//5)+(1)/((2^(x-1)+7)^(1//3)))^(8)` Now, `T_(4) = T_(3+1)=.^(8)C_(3)((4^(x)+44)^(1//5))^(8-3)(1)/((2^(x-1)+7)^(1//3))^(3)` Given `336 = .^(8)C_(3)((4^(x) + 44)/(2^(x-1) + 7))` Let `2^(x) = y` `rArr 336 = .^(8)C_(3)((y^(2)+44)/((y//2)+7))` or `336 = (8xx7xx6)/(3xx2xx1)((2(y^(2)+44))/(y+14))` `rArr y^(2) -3y + 2 =0` or `y = 0,2` or `y = 2` or `x= 4` |
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| 268. |
Find the fourth term in the expansion of `(1-2x)^(3//2)""dot` |
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Answer» We have, `(1-2x)^(3//2) = 1+(3)/(2) (-2x)+((3)/(2) xx 1/2)/(2!) (-2x)^(2)` `+ ((3)/(2)xx1/2(-1/2))/(3!) (-2x)^(3)+"….."` Hence , the `4^(th)` term is `x^(3)//2`. |
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| 269. |
If the fourth term of `(1/(x^(1+(log)_(10)x)+x 12))^6`is equal to 200 and `x >1,`then `x`is equal to`10sqrt(2)`(2) 10(3) `10^4`(4) 100A. `10^(sqrt(2))`B. 10C. `10^(4)`D. none of these |
| Answer» Correct Answer - B | |
| 270. |
if the coefficient of `(2r+1)`th term and `(r+2)`th term in the expansion of `(1+x)^(43)` are equal then r=?A. 14B. 30C. 41D. 42 |
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Answer» Correct Answer - A N/a |
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| 271. |
Show that the term independent of x in the expansion of `(x-1/x)^(10) is -252`. |
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Answer» `T_(r+1)=(-1)^(r)*^(10) C_(r)x^((10-1))*(1/x)^(r) = (-1)^(r)*^(10)C_(r)x^((10-20)).` `Putting 10-2r =0, " we get " r=5.` ` :. (T_(6) -1)^(5) *^(10)C _(5)= - ( 10 xx 9 xx 8 xx 7 xx 6 )/(5 xx 4 xx 3 xx 2 xx 1 ) = - 252.` |
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| 272. |
Show that the middle term in the expansion of `((2x^2)/3+3/(2x)^(2))^(10) " is " 252.` |
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Answer» `T_(r+1)=.^(10)C_(r) ((2x^(2))/3)^(10-r) xx (3/(x^(2)))^(r) = .^(10)C_(r) xx 2^(10-2r) xx 3 ^(2r - 10) xx x^((20-4r)).` The expansion has 11 terms . So, middle term = 6th term. `T_(6) = T_(5+1) = .^(10)C_(5) xx 2^(0) xx 3^(0) xx x^(0) = ((10 xx 9 xx 8 xx 7 xx 6 )/( 5 xx 4 xx 3 xx 2 xx 1 )) = 252.` |
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| 273. |
The constant term in the expansion of `(log(x^(logx))-log_(x^(2))100)^(12)` is (base of lof is 10) `"_____"`. |
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Answer» Correct Answer - 495 `(log(x^(logx))-log_(x^(2))100)^(12)` `= ((logx)^(2)-(1)/(logx))^(12)` `:.` Constant term `= .^(12)C_(4)((logx)^(2))^(4)(-(1)/(logx))^(8)` `= .^(12)C_(4)` `= (12xx11xx10xx9)/(24)` `=55xx9=495` |
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| 274. |
Find the following sums : (i) `sumsum_(inej) ""^(n)C_(i).""^(n)C_(j)` , (ii) `sumsum_(0leiltjlen) ""^(n)C_(i).""^(n)C_(j)`. (iii) `sumsum_(0leiltjlen) ""^(n)C_(i).""^(n)C_(j)`. |
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Answer» (i) `underset(i ne j)(sumsum).^(n)C_(i)..^(n)C_(j)=(underset(i=0)overset(n)sumunderset(j=0)overset(n)sum.^(n)C_(i).^(n)C_(j))-underset(i=0)overset(n)sum(.^(n)C_(i))^(2)` `= (underset(i=0)overset(n)sum.^(n)C_(i))(underset(j=0)overset(n)sum.^(n)C_(j))-underset(i=0)overset(n)sum(.^(n)C_(i))^(2)` `= (2^(n))(2^(n))-.^(2n)C_(n) = 4^(n)-.^(2n)C_(n)` (ii) `underset(0leiltjlen)(sumsum).^(n)C_(i)..^(n)C_(j)= ((underset(i=0)(sum)underset(j=0)(sum).^(n)C_(i).^(n)C_(j))-underset(i=0)(sum)(.^(n)C_(i))^(2))/(2)` `= (2^(2n)-.^(2n)C_(n))/(2)` (ii) `underset(0leiltjlen)(sumsum).^(n)C_(i)..^(n)C_(j)= ((underset(i=0)(sum)underset(j=0)(sum).^(n)C_(i).^(n)C_(j))+underset(i=0)overset(n)sum(.^(n)C_(i))^(2))/(2)` `= (2^(2n) + .^(2n)C_(n))/(2)` |
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| 275. |
Find the value of `sumsum_(0leiltjlen) (""^(n)C_(i)+""^(n)C_(j))`. |
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Answer» `underset(0leiltjlen)(sumsum)(.^(n)C_(i)+.^(n)C_(j))=((underset(i=0)overset(n)sumunderset(j=0)overset(n)sum(.^(n)C_(i)+.^(n)C_(j)))-underset(i=0)overset(n)sum2.^(n)C_(i))/2` `= ((underset(i=0)overset(n)sum(underset(j=0)overset(n)sum.^(n)C_(i)+underset(j=0)overset(n)sum.^(n)C_(j)))-2xx2^(n))/(2)` ` = ((underset(i=0)overset(n)sum(.^(n)C_(i)underset(j=0)overset(n)sum1+2^(n)))-2^(n+1))/(2)` `= ((underset(i=0)overset(n)sum(.^(n)C_(i)(n+1)+2^(n)))-2^(n+1))/(2)` `= ((n+1)underset(i=1)overset(n)sum.^(n)C_(i)+2^(n)underset(i=0)overset(n)sum1-2^(n+1))/(2)` ` = ((n+1)2^(n)+2^(n)(n+1)-2^(n+1))/(2)` ` = (n+1)2^(n) - 2^(n) = n2^(n)` |
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| 276. |
The value of `sum_(r=1)^(n+1) (sum_(k=1)^(n) ""^(k)C_(r-1))` (where r,k, n in N`) is equal toA. `2^(n+1)-2`B. `2^(n+1)-1`C. `2^(n+1)`D. none of these |
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Answer» Correct Answer - A `underset(r=1)overset(n+1)sum(underset(k=1)overset(n)sum.^(k)C_(r-1))= underset(r=1)overset(n+1)sum(underset(k=1)overset(n)sum(.^(k+1)C_(r)-.^(k)C_(r)))` `= underset(r=1)overset(n+1)sum(.^(n+1)C_(r)+.^(1)C_(r))` `= 2^(n+1) - 2` |
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| 277. |
The sum `sumsum_(0leilejle10) (""^(10)C_(j))(""^(j)C_(r-1))` is equal toA. `2^(10) - 1`B. `2^(10)`C. `3^(10) - 1`D. `3^(10)` |
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Answer» Correct Answer - C `underset(0leilejle10)(sum)(.^(10)C_(j))(.^(j)C_(i))` `= .^(10)C_(1)(.^(1)C_(0) + .^(1)C_(1))+.^(10)C_(2)(.^(2)C_(0)+.^(2)C_(1) + .^(2)C_(2))+"...."+.^(10)C_(10)(.^(10)C_(0) +.^(10)C_(1)+"..."+.^(10)C_(10))` `= .^(10)C_(1)(2) + .^(10)C_(2)(2^(2))+.^(10)C_(3)(2^(3))+"....."+.^(10)C_(10)(2^(10))` `= (2+1)^(10) - 1` `= 3^(10) - 1` |
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| 278. |
The value of the sum `.^(1000)C_(50) + .^(999)C_(49) +.^(998)C_(48)+"….".^(950)C_(0)` isA. `.^(1001)C_(50)`B. `.^(1002)C_(951)`C. `.^(1001)C_(950)`D. `.^(1002)C_(50)` |
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Answer» Correct Answer - A `.^(1000)C_(50)+.^(999)C_(49)+.^(998)C_(48)+"...."+.^(950)C_(0)` `=` coefficient of `x^(950)` in `[(1+x)^(950) + (1+x)^(951) + "....."+(1+x)^(1000)]` `=` coefficient of `x^(950)` in `((1+x)^(950)((1+x)^(51)-1))/((1+x)-1)` `=` coefficient of `x^(950)` in `((1+x)^(1001) - (1+x)^(950))/(x)` `=` coefficient of `x^(951)` in `(1+x)^(1001)` `= .^(1001)C_(1951) = .^(1001)C_(50)` |
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| 279. |
If the fractional part of the number `(2^(403))/(15)` is `(k)/(15)`, then k is equal toA. 14B. 6C. 4D. 8 |
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Answer» Consider , `2^(403)=2^(400+3)=8*2^(400)=8*2^(400)=8*(2^(4))^(100) =8*(16)^(100)=8(1+15)^(100) =8(1+""^(100)C_(1)(15)+""^(100)C_(2)(15)^(2)+....+""^(100)C_(100)(15)^(100))` [By binomial theorem , `(1+x)^(n) =""^(n)C_(0)+""^(n)C_(1)x+""^(n)C_(2)x^(2)+...""^(n)C_(n)x^(n) , n in N ]` `=8+8(""^(100)C_(1)(15)+""^(100)C_(2)(15)^(2)+...+""^(100)C_(100)(15)^(100))` `=8+8xx 15 lambda ` where `lambda=""^(100)C_(1)+....+""^(100)C_(100)(15)^(99)in N ` `:. (2^(403))/(15)=(8+8xx 15 lamda)/(15)=8lambda+(8)/(15)` `implies {(2^(403))/(15)}=(8)/(15)` (where `{*}` is the fractional part function ) `:. k=8` Alternate Method `2^(403)=8.2^(400)=8(16)^(100)`. Note that, when 16 is divided by 15, gives remainder 1. `therefore` When `(16)^(100)` is divided by 15, gives remainder `1^(100)=1` and when `8(16)^(100)` is divided by 15, gives remainder 8. `therefore {(2^(403))/(15)}=(8)/(15)`. (where `{.}` is the fractional part function) `rArr k=8 |
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| 280. |
Using binomial theorem, prove that `(2^(3n)-7n-1)` is divisible by 49, where n `in` N. |
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Answer» `( 1 + x ) ^ ( n ) = . ^ ( n) C _ ( 0 ) + .^ (n) C _ ( 1 ) x + .^ ( n ) C _ ( 2 ) x ^ ( 2 ) + ...+ .^ ( n ) C _ ( n)x ^ ( n ) .` `( 1 + 7 ) ^ ( n ) = . ^ ( n) C _ ( 0 ) + . ^ ( n) C_ ( 1 ) xx 7 + . ^ ( n ) C _( 2 ) xx ( 7 ) ^ ( 2 ) + ...+ . ^ ( n ) C _ ( n ) xx ( 7 ) ^ ( n ) ` `= 1 + 7 n + ( 7 ) ^ ( 2 ) [.^ ( n) C_ ( 2 ) + .^ ( n) C_ ( 3 ) xx 7 + ... + ( 7 ) ^ ( n-2 ) ]` `rArr ( 8 ^ ( n ) - 7n _ 1 ) = ( 7 ) ^ ( 2 ) xx " (an integar)" = 49 xx "( an integar) ".` , Hence, `(2 ^ ( 3n ) -7n-1)` is divisible by 49. |
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| 281. |
Prove that `(.^(n)C_(1)sin2x+.^(n)C_(2)sin4x+.^(n)C_(3)sin6x+"…..")/(1+.^(n)C_(1)cos2x+.^(n)C_(2)cos4x+.^(n)C_(3)cos 6x+"……")` |
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Answer» L.H.S. `=(underset(r=0)overset(n)sum.^(n)C_(r)sin2rx)/(underset(r=0)overset(n)sum.^(n)C_(r)cos2x)" "(1)` `0=(underset(r=0)overset(n)sum.^(n)C_(n-r)sin2(n-r)x)/(underset(r=0)overset(n)sum.^(n)C_(n-r)cos2(n-r)x)` `=(underset(r=0)overset(n)sum.^(n)C_(r)sin(2n-2r)x)/(underset(r=0)overset(n)sum.^(n)C_(r)cos(2n-2r)x)" "(2)` `=(underset(r=0)overset(n)sum.^(n )C_(r)sin2rx+underset(r=0)overset(n)sum.^(n)C_(r)sin(2n-2r)x)/(underset(r=0)overset(n)sum.^(n)C_(r)cos2rx+underset(r=0)overset(n)sum.^(n)C_(r)cos(2n-2r)x)`. (Using `a/b = c/d = (a+c)/(b+d)` and using (1) and (2)) `= (underset(r=0)overset(n)sum.^(n)C_(r)[sin2rx+sin(2n-2r)x])/(underset(r=0)overset(n)sum.^(n)C_(r)[cos2rx+cos(2n-2r)x])` `= (2sin nx underset(r=0)overset(n)sum.^(n)C_(r)cos(2r-n)x)/(2cos nx underset(r=0)overset(n)sum.^(n)C_(r)cos(2r-n)x) = tan nx` |
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| 282. |
Find the 9th term in the expansion of `(a/b-b/(2a)^(2))^(12)`. |
| Answer» Correct Answer - `(495b^(4))/(256a^(12))` | |
| 283. |
Evalute: `(2+sqrt(3))^(7)+(2-sqrt(3))^(7)` |
| Answer» Correct Answer - 110084 | |
| 284. |
Find the 7th term in the expansion of `((4x)/5+5/(2x))^(8)` |
| Answer» Correct Answer - `4375/x^(4)` | |
| 285. |
If the 7th terms from the beginning and end in the expansion of `( root(3) 2+1/(root(3)2))^(n)` are equal, find the value of n. |
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Answer» Correct Answer - n=12 ` " 7th term from the end " =(n-7+2)th " term " =(n-5) th " term ".` `:. T _ (7) = T _ ( n-5 ) rArr n-5 = 7 rArr n = 12.` |
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| 286. |
Find the 7th term in the expansion of `((4x)/5-5/(2x))^9`. |
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Answer» `T_7=T_(6+1)=9C_6((4x)/5)^(9-6)((-5)/(2x))^6` `(9!)/((6!3!))*(4)^3/(5^3)*x^3*(-5)^6/2^6*1/x^6` `(84*(5^3))/x^3` `(84*125)/x^3` `10500/x^3`. |
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| 287. |
In the expansion of `(1+x)^n ,`7th and 8th termsare equal. Find the value of `(7//x+6)^2`. |
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Answer» Correct Answer - `n^(2)` Since the `7^(th)` and `8^(th)` terms are equal , we have `.^(n)C_(6)x^(6) =.^(n)C_(7)x^(7)` or `x=(.^(n)C_(6))/(.^(n)C_(7))=(7)/(n-6)` or `(7/x+6)=n^(2)` |
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| 288. |
The coefficient of `x^(8) y^(6) z^(4)` in the expansion of `(x + y + z)^(18)`, is not equal toA. `""^(18)C_(14) xx""^(14)C_(8)`B. `""^(18)C_(10)xx""^(10)C_(6)`C. `""^(18)C_(6) xx""^(12)C_(8)`D. `""^(18)C_(6)xx""^(14)C_(6)` |
| Answer» Correct Answer - D | |
| 289. |
If `(1+x-2x^2)^6=1+a_1x+a_2x^(12)++a_(12)x^(12),`then find the value of `a_2+a_4+a_6++a_(12)dot` |
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Answer» Correct Answer - 31 `(1+x-2x^(2))^(6) = 1+a_(1)x+a_(2)x^(2)+"…."+a_(12)x^(12)` Putting `x = 1` and `x = - 1` and adding the results, we get `64 = 2(1+a_(2)+a_(4)+"….")` `:. a_(2)+a_(4)+a_(6)+"…."a_(12) = 31` |
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| 290. |
If `(1+2x+3x^2)^(10)=a_0+a_1x+a_2x^2++a_(20)x^(20),t h e na_1`equals`10`b. `20`c. `210`d. none of theseA. 10B. 20C. 210D. none of these |
| Answer» Correct Answer - B | |
| 291. |
If `(1+x+x^2)^n=a_0+a_1x+a_2x^2+....+a_(2n)x^(2n)`, then `a_0+a_2+a_4+.....+a_(2n)` isA. `(3^(n) +1)/(2)`B. `(3^(n) -1)/(2)`C. `(3^(n -1) +1).(2)`D. `(3 ^(n-1) -1)/(2)` |
| Answer» Correct Answer - a | |
| 292. |
If `(1 +x+x^2)^25 = a_0 + a_1x+ a_2x^2 +..... + a_50.x^50` then `a_0 + a_2 + a_4 + ... + a_50` is :A. evenB. odd and of the form `3n`C. odd and of the form `(3n-1)`D. odd and of the form `(3n+1)` |
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Answer» Correct Answer - A `(a)` Putting `x=1` and `-1` and adding `a_(0)+a_(2)+…+a_(50)=(3^(25)+1)/(2)` `=((1+2)^(25)+1)/(2)` `=("^(25)C_(0)+^(25)C_(1)*2+^(25)C_(2)*2^(2)+^(25)C_(25)*2^(25)+1)/(2)` `=(2[1+^(25)C_(1)+^(25)C_(2)*2+...+^(25)C_(25)*2^(24)])/(2)` `=2[13+^(25)C_(2)+...+^(25)C_(25)*2^(23)]` |
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| 293. |
If n is a positive integer, then `(sqrt(3)+1)^(2n)-(sqrt(3)-1)^(2n)`is(1) an irrational number(2) an odd positive integer(3) an even positive integer (4) a rational number other thanpositive integersA. an irrational numberB. an odd positive integerC. an even positive integerD. a rational number other than positiveintegers |
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Answer» Correct Answer - A `(3sqrt(3) + 1)^(2n)- (sqrt(3) - 1)^(2n)` `=2[.^(2n)C_(1)(sqrt(3))^(2n-1)+.^(2n)C_(3)(sqrt(3))^(2n-3)+.^(2n)C_(5)(sqrt(3))^(2n-5) + "….."]` = an irrational number |
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| 294. |
For and odd integer `n ge 1, n^(3) - (n - 1)^(3) ` + …… `+ (- 1)^(n-1) 1^(3)` |
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Answer» Since, n is an odd interger, `(-1)^(n-1)=1` `and n-1,n-3,n-5` etc., are even intergers, then `n^(3)-(n-1)^(3)+(n-2)^(3)-(n-3)^(3)+ . . .+(-1)^(n-1).1^3` `=n^(3)+(n-1)^(3)+(n-2)^(3)+ . .+1^(3)` `-2[(n-1)^(3)+(n-3)^(3)+ . . .+2^(3)]` `=sumn^(3)-2xx2^(3)[((n-1)/(2))^(3)+((n-3)/(2))^(3)+ . . .+1^(3)]` `[beausen-1,n-3` . . ., are even intergers] `=sumn^(3)-16[sum((n-1)/(2))^(3)]` `=[(n(n+1)/(2)]^(2)-16[(1)/(2)((n-1)/(2))((n-1)/(2)+1)]^(2)` `=(1)/(4)(n+1)^(2)-(16(n-1)^(2)(n+1)^(2))/(4xx4xx4)` `=(1)/(4)(n+1)^(2)[n^(2)-(n-1)^(2)]=(1)/(4)(n+1)^(2)(2n-1)` |
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| 295. |
If m,n,r are positive integers such that r `lt ` m,n, then `""^(m)C_(r)+""^(m)C_(r-1)""^(n)C_(1)+""^(m)C_(r-2)""^(n)C_(2)+...+ ""^(m)C_(1)""^(n)C_(r-1)+""^(n)C_(r)` equalsA. `(""^(n)C_(r))^(2)`B. `""^(m+n)C_(r)`C. `""^(m+n)C_(r) + ""^(m)C_(r) + ""^(n)C_(r)`D. none of these |
| Answer» Correct Answer - B | |
| 296. |
If `sum_(r=0)^(2n) a_(r) (x-2)^(r) = sum_(r=0)^(2n) b_(r),(x-3)^(r)` and `a_(k) = 1` for all `k ge n`, then `b_(n)` is equal toA. `.^(2n+1)C_(n-1)`B. `.^(2n)C_(n+1)`C. `.^(2n)C_(n)`D. `.^(2n+1)C_(n+1)` |
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Answer» Correct Answer - D In the given equation, put `x - 3 = y`. `:. underset(r=0)overset(2n)suma_(r)(1+y)^(r) = underset(r=0)overset(2n)sumb_(r)(y)^(r)` `rArr a_(0) + a_(1)(1+y) + "….."+ a_(n+1)(1+y)^(n-1)+(1+y)^(n) + (1+y)^(n+1)+"…."+(1+y)^(2n)` `= underset(r=0)overset(2n)sumb_(r)y^(r)` [Using `a_(k) = 1, AA k ge n`] Equating the coeficients of `y^(n)` on both sides, we get `.^(n)C_(n) + .^(n+1)C_(n) + .^(n+2)C_(n) + "......" + .^(2n)C_(n) = b_(n)` `rArr (.^(n+1)C_(n+1)+.^(n+1)C_(n)) + .^(n+2)C_(n) + "...." + .^(2n)C_(n) = b_(n)` [Using `.^(n)C_(n) = .^(n+1)C_(n+1) = 1`] `rArr b_(n) = .^(n+2)C_(n+1) + .^(n+2)C_(n) + "...." + .^(2n)C_(n)` Combing the term in similar way, we get `rArr b_(n) =.^(2n)C_(n+1) + .^(2n)C_(n) = .^(2n+1)C_(n+1)`. |
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| 297. |
Find the number of terms which are free from radical signs in theexpansion of `(y^(1//5)+x^(1//10))^(55)dot` |
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Answer» In the expansion of `(y^(1//5)+x^(1//10))^(55)`, `T_(r+1) = .^(55)C_(r)(y^(1//5))^(55-r)(x^(1//10))^(r)=.^(55)C_(r)y^(11-r//5)x^(r//10)` Thus, `T_(r+1)` will be independent of radicals if the exponents `r//5` and `r//10` are integers for `0 le r le 55`, which is possible only when `r = 0`, `10, 20, 30 , 40, 50`. Therefore, there are six terms, i.e., `T_(1), T_(11), T_(21), T_(31), T_(41), T_(51)` which are independent of radicals. |
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| 298. |
If the coefficients of `(r-5)^(t h)`and `(2r-1)^(t h)`terms in the expansion of `(1+x)^(34)`are equal, find `rdot` |
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Answer» Correct Answer - `r = 14` The coefficient of `(r-5)` th and `(2r-1)` the terms of the expansion `(1+x)^(34)` are `.^(34)C_(r-6)` and `.^(34)C_(2r-2)`, respectively. Since they are equal so `.^(34)C_(r-6) = .^(34)C_(2r-2)` Therefore, either `r - 6 = 2r-2` or `r - 6 = 34 - (2r-2)` [Using the fact that if `.^(n)C_(r) = .^(n)C_(p)`, then either `r = p` or `r = n - p` ] So, we get `r = - 4` or `r = 14`. r being a natural number, `r = - 4` is not possible. So `r = 14`. |
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| 299. |
For any positive integers m, n (with `n ge m`) If `({:(n),(m):}) = .^(n)C_(m)` Prove that `({:(n),(m):}) + ({:(n - 1),(m):}) + ({:(n - 2),(m):}) + … + ({:(m),(m):}) = ({:(n + 1),(m + 1):})` Prove that `({:(n),(m):}) + 2 ({:(n + 1),(m):}) + 3 ({:(n - 2),(m):}) + .... + (n - m + 1)` `({:(m),(m):}) = ({:(n + 2),(m + 2):})` |
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Answer» Let `S = ((n)/(m))+((n-1)/(m)) + ((n-2)/(m))+ ((n-2)/(m)) +.....+((m)/(m)) =((n+1)/(m+1)).....(i)` It is obvious that, `n ge m` Note : This question is based upon additive loop. Now ,`S =((m)/(m)) + ((m+1)/(m)) + ((m+2)/(m)) +.......+((n)/(m))` ` ={((m+1)/(m+1))+((m+1)/(m))}[because ((m)/(m)) = 1= ((m+1)/(m+1))]` `= ((m+2)/(m+1)) + ((m+2))/(m)) + ......+ ((n)/(m))" "[because""^(n)C_(r)+ ""^(n)C_(r+1) = ""^(n+1)C_(r+1)]` `=((m+2)/(m+1)) +......+((n)/(m))` `=.............` ` =((n)/(m+1))+((n)/(m)) = ((n+1)/(m+1))` which is ture ....(ii) Again, we have to prove that `((n)/(m))+2((n-1)/(m)) + 3((n-2)/(m)) +......+ (n-m+1)((m)/(m)) = ((m+2)/(m+2))` Let `S_(1) = ((n)/(m))+2((n-1)/(m)) +3((n-2)/(m)) +......+(n-m+1)((mm)/(m))` `{:(= ((n)/(m)) + ((n-1)/(m)) + ((n-2)/(m)) +...+ ((m)/(m))), ( " "+ ((n-1)/(m)) + ((n-2)/(m)) +...+ ((m)/(m))), (" "+ ((n-2)/(m)) +...+((m)/(m)) ),(" "...) , (" "+ ((m)/(m))):}}n-m + 1` rows Now, sum of the first row is `((n+1)/(m+1))` Sum of the second row is `((n)/(m+1))` Sum of the third row is `((n+1)/(m+1))`, .................... Sum of the last row is `((m)/(m)) = ((m+1)/(m+1))` Thus `S = ((n+1)/(m+1))+((n)/(m+1)) + ((n +1)/(m+1))+.......+ ((m+1)/(m+1)) = ((n+1+1)/(m+2)) = ((n+2)/(m+2))` [from Eq. (i) replacing n by n +1 and m by m + 1] |
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| 300. |
Find the number of irrational terms in the expansion of `(5^(1//6)+2^(1//8))^(100)dot` |
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Answer» Correct Answer - `97` `T_(r+1)`, the `(r+1)` th term in the expansion of `(5^(1//6) + 2^(1//8))^(100)`, is `T_(r+1) = .^(100)C_(r)(5^(1//6)).^(100-r)(2^(1//8))^(r)` As 5 and 2 are respectively prime, `T_(r+10` will be rational if `(100-r)//6` and `r//8` are both integers, i.e, `100 -r` is a multiple of 6 and ris a multiple of 8. As `0 le r le 100`, multiples of 8 up to 100 and corresponding values of 100 - r aer given by `r = 0, 8, 16, 24,"......",88,96` `100 -r = 100, 92,84,76,"......",12,4` In `100 - r` values, multiples of 6 ar `84, 60, 36` and `12`. Hence, there are just four rational terms. Therefore, the number of irrational terms is `101 - 4 = 97`. |
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