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301.

Prove that `2^k(n,0)(n,k)-2^(k-1)(n,1)(n-1,k-1)+2^(k-2)(n,2)(n-2,k-2)-.....+(-1)^k(n,k)(n-k,0)=(n,k)`.

Answer» To show that
`2^(k).""^(n)C_(0).""^(n)C_(k)-2^(k-1).""^(n)C_(1).""^(n-1)C_(k-1)+2^(k-2).""^(n)C_(2).""^(n)C_(k-2)-...+(-1)^(k)""^(n)C_(k)""^(n-k)C_(0)=""^(n)C_(k)`
Taking LHS
`2^(k).""^(n)C_(0).""^(n)C_(k)-2^(k-1).""^(n)C_(1).""^(n-1)C_(k-1) +... +(-1)^(k)""^(n)C_(k).""^(n-k)C_(0)`
`=sum_(r=0)^(k)(-1)^(r).2^(k-r).""^(n)C_(r)""^(n-r)C_(k-r)`
`= sum _(r=0)^(k)(-1)^(r)2^(k-r).(n!)/(r!(n-r)!).((n-r)!)/((k-r)!(n-k)!)`
`= sum _(r=0)^(k)(-1)^(r).2^(k-r).(n!)/(r!(n-r)!k!).(k!)/(r!(k-r)!)`
`= sum _(r=0)^(k)(-1)^(r).2^(k-r). ""^(n)C_(k).""^(n)C_(r)=2^(k).""^(n)C_(k){sum _(r=0)^(k)(-1)^(r).(1)/(2^(r))""^(k)C_(r)}`
`=2^(k).""^(n)C_(k)(1-(1)/(2))^(k)=""^(n)C_(k)=RHS `
Discussion
302.

Represent `cos 6 theta` in terms of `cos theta`.

Answer» Correct Answer - `32 cos^(6) theta - 48 cos^(4) theta +18 cos^(2) theta - 1`
`cos 6 theta = "Real part of"(cos theta + i sin theta)^(6)`
`= .^(6)C_(0)cos^(6)theta - .^(6)C_(2)cos^(4) theta sin^(2) theta + .^(6)C_(4) cos^(2)theta sin^(4) theta-.^(6)C_(6)sin^(6) theta`
`= cos^(6)theta - 15cos^(4)theta(1-cos^(2)theta)+15cos^(2)theta(1-cos^(2)theta)^(2)-(1-cos^(2)theta)^(3)`
`= cos^(6)theta - 15 cos^(4)theta(1-cos^(2)theta)+15cos^(2)theta(1-2cos^(2)theta+cos^(4)theta)-(1-3cos^(2)theta+3cos^(4)theta-cos^(6)theta)`
`= 32 cos^(6) theta - 48 cos^(4)theta + 18 cos^(2) theta - 1`
Discussion
303.

Find the value of `(sqrt(2)+1)^6-(sqrt(2)-1)^6dot`

Answer» Correct Answer - `140sqrt(2)`
`(x+a)^(n)-(x-a)^(n) = 2[.^(n)C_(1)x^(n-1)a+.^(n)C_(3)x^(n-3)a^(3)+.^(n)C_(5)x^(n-5)a^(5)+"……"]`
`:. (sqrt(2)+1)^(6)-(sqrt(2)-1)^(6)`
`= 2[.^(6)C_(1)(sqrt(2))^(5)(1)^(1)+.^(6)C_(3)(sqrt(2))^(3)+.^(6)C_(5)(sqrt(2))^(1)(1)^(5)]`
`= 2[6 xx 4 sqrt(2) + 20 xx 2 sqrt(2) + 6 sqrt(2)]`
`= 2[24sqrt(2)+40sqrt(2)+6sqrt(2)] = 140sqrt(2)`
Discussion
304.

The sum of the coefficeints of the polynominal `(1 + x - 3x^(2))^(2163)` is …….

Answer» Sum of coefficients is obtained by putting `x=1`.
i.e. `(1+1-3)^(2163)=-1`
Thus, sum of the coefficients of the polynomial `(1+x-3x^2)^(2163) ` is `-1`.
Discussion
305.

The coefficient of `x^(n)` in the expansion of `((1+x)^(2))/((1 - x)^(3))`, isA. `n^(2) + 2n +1`B. `2 n^(2) + n +1`C. `2n ^(2) + 2n +1`D. `2^(n) + 2n + 2`

Answer» Correct Answer - C
Discussion
306.

For `2

Answer» `{:((n),(r)):}+2{:((n),(r-1)):}+{:((n),(r-2)):}=[{:((n),(r)):}+{:((n),(r-1)):}]+[{:((n),(r-1)):}+{:((n),(r-2)):}]={:((n+1),(r)):}+{:((n+1),(r-1)):}={:((n+2),(r)):}`
`" "[because""^(n)C_(r)+""^(n)C_(r-1)=""^(n+1)C_(r)]`
Discussion
307.

Find the sum of the series `.^(84)C_(4)+6xx.^(84)C_(5)+15xx.^(84)C_(6)+20xx.^(84)C_(7)+15xx.^(84)C_(8)+6xx.^(84)C_(9)+.^(84)C_(10)`.

Answer» Correct Answer - `.^(90)C_(10)`
`.^(84)C_(4) + 6 xx .^(84)C_(5) + 15 xx .^(84)C_(6)+20 xx .^(84)C_(7)+15 xx.^(84)C_(8)+6xx.^(84)C_(9)+.^(84)C_(10)`
`= .^(6)C_(6) xx .^(84)C_(4)+ .^(6)C_(5) xx .^(84)C_(5) + .^(6)C_(4) xx .^(84)C_(6)+ .^(6)C_(3) xx .^(84)C_(7) + .^(6)C_(2) xx .^(84)C_(8)+ .^(6)C_(1)xx.^(84)C_(9) + .^(6)C_(0) xx.^(84)C_(10)`
`=` Coefficient of `x^(10)` in `(1+x)^(6)(1+x)^(84)`
`=` Coefficient of `x^(10)` in `(1+x)^(90)`
`= .^(90)C_(10)`
Discussion
308.

Find the number of nonzero terms in the expansion of `(1+3sqrt(2)x)^9+(1-3sqrt(2)x)^9dot`

Answer» Correct Answer - 5
Given expression is
`2[1+.^(9)C_(2)(3sqrt(2)x)^(2)+.^(9)C_(4)(3sqrt(2)x)^(4)+.^(9)C_(6)(3sqrt(2)x)^(6) + .^(9)C_(8)(3sqrt(2)x)^(8)]`
Therefore, the number of nonzero terms is 5.
Discussion
309.

Find the `13^(t h)`term in the expansion of `(9x-1/(3sqrt(x)))^(18),x!=0`

Answer» Correct Answer - 18564
Discussion
310.

If `(1 + x + x^2)^n = (C_0 + C_1x + C_2 x^2 + ............)` then the value of `C_0 C_1-C_1C_2+C_2C_3.....`A. `3^(n)`B. `(-1)^(n)`C. `2^(n)`D. none of these

Answer» Correct Answer - d
Discussion
311.

If the coefficients of rth, `(r+1)t h ,a n d(r+2)t h`terms in the expansion of `(1+x)^(14)`are in A.P., then `r`is/area. 5 b. 11 c. `10`d. `9`A. 5,9B. 6,9C. 7,9D. none of these

Answer» Correct Answer - a
Discussion
312.

Let `a=3^(1/(223))+1`and for all `geq3,l e tf(n)=^n C_0dota^(n-1)-^n C_1dota^(n-2)+^n C_2dota^(n-3)-+(-1)^(n-1)dot^n C_(n-1)dota^0`. If the value of `f(2007)+f(2008)=3^k w h e r ek in N ,`then the value of `k`is.

Answer» Correct Answer - 9
`f(n)=.^(n)C_(0)a^(n-1)-.^(n)C_(1)a^(n-2)+.^(n)C_(2)a^(n-3)+"...."+(-1)^(n-1).^(n)C_(n-1)a^(0)`
`= 1/a(.^(n)C_(0)a^(n)-.^(n)C_(1)a^(n-1)+.^(n)C_(2)a^(n-2)+"...."+(-1)^(n-1)C_(n-1)a)`
`=1/a((a-1)^(n)-(-1)^(n).^(n)C_(n))`
`= 1/a((3^(n/223)-(-1)^(n)))`
`f(x) = (3^(n/223)-(-1)^(n))/((3^(1/223)+1))`
`rArr f(2007)= (3^(2007/223)+1)/(3^(1/223)+1)`
`rArr f(2008)= (3^(2008/223)+1)/(3^(1/223)+1)`
`f(2007)+f(2008)= (3^(2007/223)+3^(2008/223))/(3^(1/223+1))`
`= (3^(9)+3^(9+(1)/(223)))/(3^(1/223)+1)`
`= 3^(9)((1+3^(1/223)))/((1+3^(1/223)))=3^(9)`
`rArr 3^(9)= 3^(k)` then `k = 9`
Discussion
313.

If `(1+x)^(15)=C_0+C_1x+C_2x^2++C_(15)x^(15),`then find the su of `C_1+2C_3+3C_4++14 C_(15)dot`

Answer» Correct Answer - `13 xx 12^(14) + 1`
`C_(2)+2C_(3)+3C_(4)+"……."+14C_(15)`
`= underset(r=1)overset(14)sumr.^(15)C_(r+1)`
`= underset(r=1)overset(14)sum[(r+1)-1].^(15)C_(r+1)`
`= underset(r=1)overset(14)sum[(r+1)^(15)C_(r+1)-.^(15)C_(r+1)]`
`=underset(r=1)overset(14)sum(15.^(14)C_(r)-.^(15)C_(r+1))`
`= 15(2^(14)-1)-(2^(15)-.^(15)C_(0) - .^(15)C_(1))`
` = 13 xx 2^(14) + 1`
Discussion
314.

If `a_(n) =sum_(r=0)^(n) (1)/(""^(n)C_(r ))`, then `sum_(r=0)^(n) (r )/(""^(n)C_(r ))` equalsA. `(n - 1) a_(n)`B. `n a_(n)`C. `(1)/(2) n a_(n)`D. None of these

Answer» Let `b=sum _(r=0)^(n)(r)/(.^Nc_r)=sum_(r=0)^(n)(-(n-r))/(.^Nc_R)`
`=n sum_(r=0)^(n)(1)/(.^n C_r)-sum_(r=0)^(n)(n-r)/(.^C_r)`
`=na_(n)-sum_(r=0)^(n) (n-r)/(.^nC_(n-r))[because .^Nc_r=.^nC_(n-r)]`
`=na_(n)-brArr 2b=na_(n)rArr b=(n)/(2)a_(n)`.
Discussion
315.

The coefficient of `x^(n)` in the expansion of `(1)/((1-x)(3 -x))`, isA. `(3^(n+1)-1)/(2.3^(n+1))`B. `(3^(n+1)-1)/(3^(n+1))`C. `2((3^(n+1)-1)/(3^(n+1)))`D. none of these

Answer» Correct Answer - A
Discussion
316.

Let `X=( ^(10)C_1)^2+2( ^(10)C_2)^2+3( ^(10)C_3)^2+ ddot +10( ^(10)C_(10))^2`, where ` ^(10)C_r`, `r in {1, 2, ddot, 10}`denote binomial coefficients. Then, the value of `1/(1430) X`is _________.

Answer» We have,
`X=(""^(10)C_(1))^(2)+2(""^(10)C_(2))^(2)+3(""^(10)C_(3))^(2)+...+10(""^(10)C_(10))^(2)`
`impliesX=underset(r=1)overset(10)sumr(""^(10)C_(r))^(2)impliesX=underset(r=1)overset(10)sumr^(10)C_(r)""^(10)C_(r)`
`impliesX=underset(r=1)overset(10)sumrxx(10)/(r)""^(9)C_(r-1)""^(10)C_(r)" "[because""^(n)C_(r)=(n)/(r)""^(n-1)C_(r-1)]`
`impliesX=10underset(r=1)overset(10)sum""^(9)C_(r-1)""^(10)C_(r)`
`impliesX=10underset(r=1)overset(10)sum""^(9)C_(r-1)""^(10)C_(r)" "[because""^(n)C_(r)=""^(n)C_(n-r)]`
`impliesX=10xx""^(19)C_(9)" "[because""^(n-1)C_(r-1)""^(n)C_(n-r)=""^(2n-1)C_(n-1)]`
Now, `(1)/(1430)X=(10xx""^(19)C_(9))/(1430)=(""^(19)C_(9))/(148)=(""^(19)C_(9))/(11xx13)`
`=(19xx17xx16)/(8)=19xx34=646`
Discussion
317.

The coefficient of `x^6` in the expansion of `(1+x+x^2)^(-3),` isA. 6B. 5C. 4D. 3

Answer» Correct Answer - D
Discussion
318.

The coefficient of `x^(10)`in the expansion of `(1+x^2-x^3)^8`is`476`b. `496`c. `506`d. `528`A. 476B. 496C. 506D. 528

Answer» Correct Answer - A
Discussion
319.

Evaluate`.^(n)C_(0).^(n)C_(2)+2.^(n)C_(1).^(n)C_(3)+3.^(n)C_(2).^(n)C_(4)+"...."+(n-1).^(n)C_(n-2).^(n)C_(n)`.

Answer» Correct Answer - `n..^(2n-1)C_(n-3)+.^(2n)C_(n-2)`
`S = .^(n)C_(0).^(n)C_(2)+2.^(n)C_(1).^(n)C_(3)+3.^(n)C_(2).^(n)C_(4)+"....."+(n-1).^(n)C_(n-2).^(n)C_(n)`
` = .^(n)C_(0).^(n)C_(n-2)+2.^(n)C_(1).^(n)C_(n-3)+3.^(n)C_(2).^(n)C_(n-4)+"....."(n-1).^(n)C_(n-2).^(n)C_(0)`
`= underset(r=1)overset(n)sumr.^(n)C_(r-1).^(n)C_(n-r-1)`
`= underset(r=1)overset(n)sum((r-1)+1).^(n)C_(r-1).^(n)C_(n-r-1)`
`=underset(r=1)overset(n)sum[(r-1).^(n)C_(r-1).^(n)C_(n-r-1)+.^(n)C_(r-1).^(n)C_(n-r-1)]`
`= underset(r=1)overset(n)sum[n^(n-1)C_(r-2).^(n)C_(n-r-1)+.^(n)C_(r1).^(n)C_(n-r-1)]`
`= n^(2n-1)C_(n-3)+.^(2n)C_(n-2)`
Discussion
320.

Find the number of nonzero terms in the expansion of `(1+3sqrt(2)x)^9+(1-3sqrt(2)x)^9dot`A. 9B. 0C. 5D. 10

Answer» Correct Answer - C
Discussion
321.

If `C_r` stands for `nC_r,` then the sum of first `(n+1)` terms of the series `a C_0-(a+d)C_1+(a+2d)C_2-(a+3d)C_3+......,` isA. `(a)/(2^(n))`B. `na`C. 0D. none of these

Answer» Correct Answer - c
Discussion
322.

If `^n+1C_(r+1)dot^n C_rdot^(n-1)C_(r-1)=11 :6:3,`then `n r=``20`b. `30`c. `40`d. `50`A. `20`B. `30`C. `40`D. `50`

Answer» Correct Answer - D
Given,
`(.^(n+1)C_(r+1))/(.^(n)C_(r))= 11/6` or `((n+1)/(r+1)xx.^(n)C_(r))/(.^(n)C_(r)) = 11/6`
or `6n+6 = 11 r + 11` or `6n-11r = 5" "(1)`
Also,
`(.^(n)C_(r))/(.^(n-1)C_(r-1)) = 6/3` or `(n/rxx.^(n-1)C_(r-1))/(.^(n-1)C_(r-1)) = 6/3` or `n = 2r " " (2)`
From (1) and (2), `r = 5` and `n = 10`,
`:. nr = 50`
Discussion
323.

Find the value of `.^(20)C_(0) xx .^(13)C_(10) - .^(20)C_(1) xx .^(12)C_(9) + .^(20)C_(2) xx .^(11)C_(8) - "……" + .^(20)C_(10)`.

Answer» `.^(20)C_(0) xx .^(13)C_(10) - .^(20)C_(1) xx .^(12)C_(9) + .^(20)C_(2) xx .^(11)C_(8) -"…." + .^(20)C_(10)`
`=` Coefficient of `x^(10)` in
`(.^(20)C_(0) - .^(20)C_(1)x + .^(20)C_(2)x^(2) - .^(20)C_(3)x^(3) + "…..")`
`(1+.^(4)C_(1)x + .^(5)C_(2)x^(2)+.^(6)C_(3)x^(3)+".....")`
`=` Coefficient of `x^(10)` in `(1-x)^(20)(1-x)^(-4)`
`=` Coefficient of `x^(10)` in `(1-x)^(16)`
`= .^(16)C_(10)`.
Discussion
324.

If `C_r` stands for `nC_r`, then the sum of the series `(2(n/2)!(n/2)!)/(n !)[C_0^2-2C_1^2+3C_2^2-........+(-1)^n(n+1)C_n^2]` ,where n is an even positive integer, is

Answer» Correct Answer - C
Discussion
325.

If `C_r` stands for `nC_r`, then the sum of the series `(2(n/2)!(n/2)!)/(n !)[C_0^2-2C_1^2+3C_2^2-........+(-1)^n(n+1)C_n^2]` ,where n is an even positive integer, isA. `(-1)^(n//2) (n + 2)`B. `(-1)^(n) (n + 1)`C. `(-1)^(n//2) (n + 1)`D. None of these

Answer» We have, `C_(0)^(2)-2C_(1)^(2)+3C_(2)^(2)-4C_(3)^(2)+...+(-1)^(n)(n+1)C_(n)^(2)`
`=[C_(0)^(2)-C_(1)^(2)+C_(2)^(2)-C_(3)^(2)+...+(-1)^(n)C_(n)^(2)]+[C_(1)^(2)-2C_(2)^(2)+3C_(3)^(2)-...+(-1)^(n)nC_(n)^(2)]`
`=(-1)^(n//2)(n!)/(((n)/(2))!((n)/(2))!)-(-1)^((n)/(2)-1)(n)/(2)(n!)/(((n)/(2))!((n)/(2))!)`
`=(-1)^(n//2)(n!)/(((n)/(2))!((n)/(2))!)(1+(n)/(2))`
`:.(2((n)/(2))!((n)/(2))!)/(n!)[C_(0)^(2)-2C_(1)^(2)+3C_(2)^(2)-...+(-1)^(r)(n+1)C_(n)^(2)]`
`=(2((n)/(2))!((n)/(2))!)/(n!)(-1)^(n//2)(n!)/(((n)/(2))!((n)/(2))!)((n+2))/(2)=(-1)^(n//2)(n+2)`
Discussion
326.

Prove that `sum_(r=0)^(n) r(n-r)C_(r)^(2)=n^(2)(""^(2n-2)C_(n))`.

Answer» `underset(r=0)overset(n)sumr(n-r)(.^(n)C_(r))^(2) = underset(r=0)overset(n)sumr.^(n)C_(r)(n-r).^(n)C_(n-r)`
`=underset(r=0)overset(n)sumn.^(n-)C_(r-1)n.^(n-1)C_(n-r-1)`
`= n^(2)underset(r=0)overset(n)sum.^(n-1)C_(r-1).^(n-1)C_(n-r-1)`
`= n^(2) xx "coefficient of" x^(n-2) "in" (1+x)^(n-1)(1+x)^(n-1)`
`= n^(2) xx .^(2n-2)C_(n-2) = n^(2).^(2n-2)C_(n)`
Discussion
327.

Prove that `(.^(n)C_(1))/(2) + (.^(n)C_(3))/(4) + (.^(n)C_(5))/(6) + "…." = (2^(n) - 1)/(n+1)`.

Answer» `S = (.^(n)C_(1))/(2)+(.^(n)C_(3))/(4)+(.^(n)C_(5))/(6)+"..."`
`:. T_(r) = (.^(n)C_(2r-1))/(2r)= (.^(n+1)C_(2r))/(n+1)`, where `r = 1,2,3,"...."`
`rArr S = 1/(n+1)(.^(n+1)C_(2)+.^(n+1)C_(4)+.^(n+1)C_(6)+"....")`
` = (1)/(n+1)[(.^(n+1)C_(0) + .^(n+1)C_(2) + .^(n+1)C_(4) +"....")-.^(n+1)C_(0)] = (2^(n)-1)/(n+1)`.
Discussion
328.

In the expansion of `(1+x)^(50),`find the sum of coefficients of odd powers of `xdot`

Answer» Correct Answer - `2^(49)`
We have,
`(1+x)^(50) = .^(r=0)overset(5)sum.^(50)C_(r )x^(r )`
Therefore, sum of coefficient of odd powers of x is
`.^(50)C_(1) + .^(50)C_(3) + "……" + .^(50)C_(49) = (1)/(2) (2^(50)) = 2^(49)`
Discussion
329.

The sum of all the coefficients of the terms in the expansion of `(x+y+z+w)^(6)` which contain `x` but not `y`, isA. `3^(6)`B. `2^(6)`C. `3^(6)-2^(6)`D. none of these

Answer» Correct Answer - C
`(c )` The sum of coefficients of terms not containing `y=3^(6)` (putting `x=z=w=1` and `y=0`)
The sum of coefficients of term not containing both `x` and `y=2^(6)` (putting `z=w=1` and `x=y=0`)
So the required number `=3^(6)-2^(6)`
Discussion
330.

The coefficient of y in the expansion of `(y^(2) + c//y)^(5)` isA. 29cB. 10 cC. `10c^(3)`D. `20 c^(2)`

Answer» Correct Answer - C
Discussion
331.

Find the sum `2..^(10)C_(0) + (2^(2))/(2).^(10)C_(1) + (2^(3))/(3).^(10)C_(2)+(2^(4))/(4).^(10)C_(3)+"...."+(2^(11))/(11).^(10)C_(10)`.

Answer» `2..^(10)C_(0)+(2^(2))/(2) .^(10)C_(1) + (2^(3))/(3) .^(10)C_(2)+(2^(4))/(4).^(10)C_(3) + "...." + (2^(11))/(11).^(10)C_(10)`
` = underset(r=0)overset(10)sum(.^(10)C_(r))/(r+1)2^(r+1)=underset(r=0)overset(10)sum(.^(11)C_(r+1)2^(r+1))/(11)`
` = ((1+2)^(11)-1)/(11) = (3^(11)-1)/(11)`
Discussion
332.

Prove that `1/(n+1)=(^n C_1)/2-(2(^n C_2))/3+(3(^n C_3))/4-+(-1)^(n+1)(n(^n C_n))/(n+1)`.

Answer» `S = (.^(n)C_(1))/(2)-(2(.^(n)C_(3)))/(3)+(3(.^(n)C_(3)))/(4)"...."+(-1)^(n+1)(n(.^(n)C_(n)))/(n+1)`
` = underset(r=1)overset(n)sum (r.^(n)C_(r))/((r+1))(-1)^(r+1)`
`= underset(r=1)overset(n)sum r(.^(n+1)C(r+1))/(n+1)(-1)^(r+1)`
`= 1/(n+1) underset(r=1)overset(n)sum [(r+1).^(n+1)C_(r+1)(-1)^(r+1)-.^(n+1)C_(r+1)(-1)^(r+1)]`
`= (1)/(n+1)underset(r=1)overset(n)sum[-(n+1).^(n)C_(r)(-1)^(r)-.^(n+1)C_(r+1)(-1)^(r+1)]`
`= -[-.^(n)C_(2)-.^(n)C_(3)+"...."+(-1)^(n).^(n)C_(n)]`
` - 1/(n+1)[.^(n+1)C_(2) - .^(n+1)C_(3)+".....(-1)^(n+1).^(n+1)C_(n+1)]`
`= - [(1-1)^(n)-1]-(1)/(n+1)[(1-1)^(n+1) - .^(n+1)C_(0)+.^(n+1)C_(1)]`
` = 1 - (1)/(n+1)[(n+1)-1]`
`1 - (n)/(n+1)= (1)/(n+1)`
Discussion
333.

Prove that `C_0 – 2^2 C_1 + 3² C_2 – 4^2 C_3 + ... +(-1)^n (n + 1)^2 C_n = 0` where `C_r = nC_r`

Answer» `C_(0) - 2^(2) C_(1) + 3^(2) . C_(2) - … + (-1)^(n) (n+1)^(2) . C_(n)`
`= sum_(r = 0 )^(n) *(-1)^(r) (r + 1)^(2)""^(n)C_(r) = sum_(r= 0) ^(n) (-1)^(r)(r^(2) + 2r + 1 ) ""^(n)C_(r)`
`= sum_(r = 0 )^(n) *(-1)^(r)r^(2). ""^(n)C_(r) +2 sum_(r= 0) ^(n) (-1)^(r)r. ""^(n)C_(r) + sum_(r=0)^(n) (-1)^(r). ""^(n)C_(r)`
`= sum_(r = 0 )^(n) *(-1)^(r)r(r - 1) ""^(n)C_(r) +3 sum_(r= 0) ^(n) (-1)^(r)r. ""^(n)C_(r) + sum_(r=0)^(n) (-1)^(r). ""^(n)C_(r)`
`= sum_(r = 0 )^(n) *(-1)^(r)n(n - 1) ""^(n - 2)C_(r-2) +3 sum_(r= 0) ^(n) (-1)^(r)n. ""^(n-1)C_(r-1) + sum_(r=0)^(n) (-1)^(r). ""^(n)C_(r)`
` = n(n- 1) {""^(n-2)C_(0) - ""^(n - 2)C_(0) - ""^(n-2)C_(1) + ""^(n-2)C_(2^(-))...+(-1)^(n) ""^(n-2)C_(n-2)}`
` = + 3n{-""^(n-1)C_(0)+ ""^(n - 1)C_(1) - ""^(n-1)C_(2)+ ...+(-1)^(n) ""^(n-1)C_(n-1)} + {""^(n)C_(0) -""^(n)C_(1) + ""^(n)C_(2) +...+ (-1)^(n)""^(n)C_(n)}
`= n(n-1). 0 + 3n .0 - AA n gt 2 = 0 , AA n gt 2 ` .
Discussion
334.

Find the largest term in the expansion of `(3+2x)^(50),w h e r ex=1//5.`A. 5th, 6thB. 51stC. 6th, 7thD. 7th, 8th

Answer» Correct Answer - C
Discussion
335.

The largest coefficient in the expansion of `(1 + x)^(30)` is.......

Answer» Largest coefficient in the expansion of `(1 + x)^(30) = .^(30)C_(30//2) = .^(30)C_(15)`
Discussion
336.

Prove that ` sum_(r=1)^(n)(-1)^(r-1)(1+1/2+1/3+"....."+1/r)""^(n)C_(r) = 1/n`.

Answer» `sum(-1)^(r-1).^(n)C_(r)(1/1+1/2+1/3+"...."+1/r)`
`=sum((-1)^(r-1).^(n)C_(r)underset(0)overset(1)int(1+x+x^(2)+"....."+x^(r-1))dx)`
`=sum(-1)^(r-1)..^(n)C_(r)underset(0)overset(1)int((1-x^(r))/(1-x))dx`
`= underset(0)overset(1)intunderset(r=1)overset(n)sum((-1)^(r-1)..^(n)C_(r)-(-1)^(r-1)..^(n)C_(r)x^(r))/(1-x)dx`
`=underset(0)overset(1) int(.^(n)C_(0)+(1-x)^(n) + (1-x)^(n))/(1-x)dx`
`= underset(0)overset(1)int(1-x)^(n-1)dx`
`= [(-(1-x)^(n-1))/(n-1)]_(0)^(1)`
`= 1/n`
Discussion
337.

In the expansion of `(1+x+x^3+x^4),`the coefficient of `x^4`is`^40 C_4`b. `^10 C_4`c. `210`d. `310`A. `""^(40)C_(4)`B. `""^(10)C_(4)`C. 210D. 310

Answer» Correct Answer - D
Discussion
338.

The middle term in the expansion of `(1-3x+3x^2-x^3)^(2n)` isA. `((6n)!x^(n))/((3n)!(3n)!)`B. `((6n)!x^(3n))/((3n)!)`C. `((6n))/((3n)!(3n)!)(-x)^(3n)`D. None of these

Answer» Correct Answer - C
`(c )` `(1-3x+3x^(2)-x^(3))^(2n)=(1-x)^(6n)`
middle term is `((6n)/(2)+1)^(th)`
i.e., `(3n+1)^(th)` term is middle term,
`T_(3n+1)=^(6n)C_(3n)(-x)^(3n)=(6n!)/(3n!3n!)(-x)^(3n)`
Discussion
339.

In the expansion of `(1+x)^(70)`, the sum of coefficients of odd powers of `x` isA. `0`B. `2^(69)`C. `2^(70)`D. `2^(71)`

Answer» Correct Answer - B
`(b)` The sum of the coefficients of odd powers in the expansion of `(1+x)^(n)`
`=` sum of the coefficients of even powers in `(1+x)^(n)`
`=2^(n-1)=2^(70-1)=2^(69)`
Discussion
340.

Using binomial theorem, find the value of `(0.99)^(15)` up to four places of decimal.

Answer» We have
`(0.99)^(15)=(1-0.01)^(15)`
`=1-.^(15)C_(1) xx(0.01) + .^(15)C_(2) xx (0.01)^(2)-.^(15)C_(3) xx (0.01)^(3) +...`
[neglecting higher powers of 0.01]
`=1-15 xx(0.01) + 105 xx (0.0001)-455 xx (0.000001)+...`
`=1-0.15+0.0105 + 0.000455 = 0.860045.`
Hence, `(0.99)^(15) = 0.860045.`
Discussion
341.

The greatest coefficient in the expansion of ` (1 + x)^(10)`, isA. `(10!)/(5!6!)`B. `(10!)/((5!)^(2))`C. `(10!)/((5!7!))`D. none of these

Answer» Correct Answer - B
Discussion
342.

Let `(1+x^(2))^(2) (1+x)^(n) = sum_(k=0)^(n+4) a_(k)x^(k)`.. If `a_(1), a_(2)` and `a_(3)` aer in arithmetic progression, then the possible value/values of n is/areA. 5B. 4C. 3D. 2

Answer» Correct Answer - B::C::D
`L.H.S. = (1+2x^(2) + x^(4)) (1+ C_(1)x + C_(2)x^(2) + C_(3)x^(3) + "…..")`
`R.H.S. = a_(0) + a_(1)x + a_(2)x^(2) + a_(3)x^(3) + "….."`
Comparing the coefficients of `x, x^(2), x^(3),"….."`
`a_(1) = C_(1), a_(2) = C_(2) + 2, a_(3) = C_(3) + 2C_(1) " "(1)`
Now, `2a_(2) = a_(1) + a_(3) (A.P.)`
`rArr 2(.^(n)C_(2) +2) = .^(n)C_(1) + (.^(n)C_(3) + 2.^(n)C_(1))` [Using (1)]
or `2 (n(n-1))/(2) + 4 = 3n + (n(n-1)(n-2))/(6)`
or `n^(3) -9n^(2) + 26n - 24 = 0`
or `(n-2)(n^(2) - 7n + 12) = 0`
or `(n-2)(n-3)(n-4) = 0`
or `n = 2,3,4`
Discussion
343.

The algebraically second largest term in the expansion of `(3-2x)^(15)` at `x=(4)/(3)`.A. `5`B. `7`C. `9`D. `11`

Answer» Correct Answer - B
`(b)` `(3-2x)^(15)` at `x=(4)/(3)`
For greatest term, `r le (16|2x|)/(3+|2x|)=(16xx2xx(4)/(3))/(3+(8)/(3))=(128)/(17)`
`:.r=7`
`:.t_(r+1)=t_(8)=^(15)C_(7)3^(8)(-2x)^(7)=-^(15)C_(7)3^(8)((8)/(3))^(7)`
greates term
`:.t_(7)` and `t_(9)` are positive term
`t_(7)=^(13)C_(6)3^(9)(-2x)^(6)=^(15)C_(6)3^(9)(8//3)^(6)=^(15) C_(6)8^(6)*3^(3)`
and `t_(9)=^(15)C_(8)3^(7)(-2x)^(8)=^(15)C_(8)3^(7)(8//3)^(8)=^(15) C_(8)8^(8)*3^(-1)`
`:.t_(9)-t_(7)=(15!)/(8!7!)(8^(8))/(3)-(15!)/(8!6!)8^(6)*3^(3)=(15!)/(9!6!)(8^(6))/(3)[(64)/(7)-(3^(4))/(9)] gt 0`
`:.t_(7) lt t_(9)`
`t_(11)=^(15)C_(10)(8^(10))/(3^(5))`
`:.t_(7)-t_(11) gt 0`
`:.t_(7) gt t_(11)`
Thus `t_(9) gt t_(7) gt t_(11)`
Hence `t_(7)` is the second largest term.
Discussion
344.

Prove that `(C_1)/1-(C_2)/2+(C_3)/3-(C_4)/4++((-1)^(n-1))/n C_n=1+1/2+1/3++1/ndot`

Answer» We have,
`(1+x)^(n)=.^(n)C_(0)+.^(n)C_(1)x+.^(n)C_(2)x^(2)+"....."+.^(n)C_(n)x^(n)`
Dividing by x, we get
`((1-x)^(n)-1)/(x) = .^(n)C_(1) + .^(n)C_(2)x+"....."+.^(n)C_(n)x^(n-1)`
`:. 1+(1+x)+(1+x)^(2) + "...."+(1+x)^(n-1)`
`= .^(n)C_(1) + .^(n)C_(2)x + "..." + .^(n)C_(n)x^(n-1)`.
`:. underset(-1)overset(0)int(.^(n)C_(1) + .^(n)C_(2)x+C_(3)x^(2) + "...."+.^(n)C_(n) x^(n-1))dx`
`= underset(-1)overset(0)int[1+(1+x)+"....."+(1+x)^(n-1)]dx`
`rArr [.^(n)C_(1)x+(.^(n)C_(2)x^(2))/(2)+(.^(n)C_(3)x^(2))/(3)+"...."+(.^(n)C_(n)x^(n))/(n)]_(-1)^(0)`
`= [x+((1+x)^(2))/(2)+((1+x)^(3))/(3)+"....."+((1+x)^(n))/(n)]_(-1)^(0)`
`rArr (.^(n)C_(1))/(1)-(.^(n)C_(2))/(2)+(.^(n)C_(3))/(3)-"...."+((-1)^(n-1))/(n).^(n)C_(n)=1+1/2+1/3+"......"+1/n`
Discussion
345.

Find the remainder when `27^(40)`is divided by 12.

Answer» `27^(40) = 3^(120)`
`3^(119) = (4-1)^(119)`
` = .^(119)C_(0)4^(119) - .^(119)C_(1)4^(118) + .^(119)C_(2)4^(117) - .^(119)C_(3)4^(116)+"….."+(-1)`
`= 4k - 1`
or `3^(120) = 12k - 3`
`= 12(k-1)+9`
Therefore, the required remainder is 9.
Discussion
346.

The value of `50 sum_(r=1)^(49)(2r^(2) - 48r +1)/((50-r).""^(50)C_(r))` is `"_____"`.

Answer» Correct Answer - 2499
`underset(r=1)overset(49)sum(2r^(2)-48r+1)/((50-r)..^(50)C_(r))= underset(r=1)overset(49)sum((r+1)^2-r(50-r))/((50-r).^(50)C_(r))`
`= underset(r=1)overset(49)sum(((r+1^(2)))/((50-r)..^(50)C_(r))-(r)/(.^(50)C_(r)))`
`=underset(r=1)overset(49)sum((r+1)/((50-r)((.^(50)C_(r))/(r+1)))-(r)/(.^(50)C_(r)))`
`=underset(r=1)overset(49)sum((r+1)/(.^(50)C_(r+1))-(r)/(.^(50)C_(r)))`
`=(50)/(.^(50)C_(50))-(r)/(.^(50)C_(1))`
`= 50 - 1/50`
Discussion
347.

Let `(1 + x)^(n) = sum_(r=0)^(n) a_(r) x^(r)` . Then `( 1+ (a_(1))/(a_(0))) (1 + (a_(2))/(a_(1)))…(1 + (a_(n))/(a_(n-1)))` is equal toA. `((n+1)^(n+1))/(n!)`B. `((n +1)^(n))/(n!)`C. `(n^(n-1))/((n -1)!)`D. `((n +1)^(n-1))/((n -1)!)`

Answer» Correct Answer - B
Discussion
348.

Let `(2x^(2)+3x+4)^(10)=sum_(r=0)^(20)a_(r )x^(r )`, then the value of `(a_(7))/(a_(13))` isA. `6`B. `8`C. `12`D. `16`

Answer» Correct Answer - B
`(b)` Given `(2x^(2)+3x=4)^(10)=sum_(r=0)^(20)a_(r )x^(r )`
Replacing `x` by `(2)/(x)`, we get
`((8)/(x^(2))+(6)/(x)+4)^(10)=sum_(r=0)^(20)a_(r )((2)/(x))^(r )`
`implies2^(10)(2x^(2)+3x+4)^(10)=sum_(r=0)^(20)a_(r )2^(r )x^(20-r)`
`impliessum_(r=0)^(20)a_(r )x^(r )=sum_(r=0)^(20)a_(r )2^(r-10)x^(20-r)`
Comparing coefficient `x^(7)` both sides , we get `a_(7)=a_(13)xx2^(3)`.
Discussion
349.

If `a,b,c,d` be four consecutive coefficients in the binomial expansion of `(1+x)^(n)`, then value of the expression `(((b)/(b+c))^(2)-(ac)/((a+b)(c+d)))` (where `x gt 0` and `n in N`) isA. positiveB. negativeC. zeroD. depends on `n`

Answer» Correct Answer - A
`(a)` `a=^(n)C_(r-1)`, `b=^(n)C_(r )`, `c=^(n)C_(r+1)`, `d=^(n)C_(r+2)`
`a+b=^(n+1)C_(r )`
`b+c=^(n+1)C_(r+1)`
`c+d=^(n+1)C_(r+2)`
`(a+b)/(a)=(n+1)/(r )`
`implies(a)/(a+b)=(r )/(n+1)`, `(b)/(b+c)=(r+1)/(n+1)`, `(c )/(c+d)=(r+2)/(n+1)`
`:. (a)/(a+b)`, `(b)/(b+c)`, `(c )/(c+d)` are in `A.P.`
`A.M. gt G. M.`
`(b)/(b+c) gt sqrt((ac)/((a+b)(c+d)))`
`implies((b)/(b+c))^(2)-(ac)/((a+b)(c+d)) gt 0`
Discussion
350.

The value of `sum_(0leiltjle5) sum(""^(5)C_(j))(""^(j)C_(i))` is equal to `"_____"`

Answer» Correct Answer - 211
`underset(0leiltjle5)(sumsum)(.^(5)C_(j))(.^(j)C_(i))`
`=.^(5)C_(1)..^(1)C_(0)+.^(5)C_(2)(.^(2)C_(0)+.^(2)C_(1))+"......"+.^(5)C_(5)(.^(5)C_(0)+.^(5)C_(1)+"...."+.^(5)C_(5))`
`= .^(5)C_(1)(2^(1)-1)+.^(5)C_(2)(2^(2)-1)+"....."+.^(5)C_(5)(2^(5)-1)`
`= (.^(5)C_(1).2^(1)+.^(5)C_(2).2^(2)+".....".+.^(5)C_(5).2^(5))-(.^(5)C_(1)+.^(5)C_(2)+"...."+.^(5)C_(5))`
`= [(1+2)^(5)-1][2^(5)-1]`
`= 3^(5)-2^(5)`
`= 243-32=211`
Discussion