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501.

The sum to infinity of the series `1+2(1-(1)/(n))+3(1-(1)/(n))^(2)+ . . .. . . .`, isA. `n^(2)`B. n(n+1)C. `n(1+(1)/(n))^(2)`D. none of these

Answer» Correct Answer - A
Let `x=1-(1)/(n)` and S be the required sum. Then
`S=1+2x+3x^(2)+4x^(2)+ . . .`
`rArr" "Sx=x+2x^(2)+3x^(3)+ . . .`
`rArr" "S(1-x)=1+x+x^(2)+x^(3)+ . . .`
`rArr" "S(1-x)=(1)/(1-x)`
`rArr" "S=(1)/((1-x)^(2))rArrS=(1)/([1-(1-1//n)]^(2))=n^(2)`
Discussion
502.

Find the nth term of the series `1+5+18+58+179+"..."`.

Answer» The sequence of first consctive differences is `4,13,40,121,"…"` and second consecutive differences is `9,27,81,"….",`. Clearly, it is an GP with comon ratio 3. So, let the nth term and sum of the series upto n terms of the series be `T_(n)` and `S_(n)`, respectively. Then,
`S_(n)=1+5+18+58+179+".."+T_(n-1)+T_(n)"....(i)"`
`S_(n)=1+5+18+58+".."+T_(n-1)+T_(n)"....(ii)"`
Subtracting Eq. (ii) from Eq.(i), we get
`0=1+4+13+40+121+".."+(T_(n)-T_(n))-T_(n)`
`implies T_(n)=1+4+13+40+121+".." " upto n terms "`
or ` T_(n)=1+4+13+40+121+".."+t_(n-1)+t_(n)"....(iii)"`
` T_(n)=1+4+13+40+".."+t_(n-1)+t_(n)"....(iv)"`
Now, subtracting Eq. (iv) from Eq. (iii), we get ` 0=1+3+9+27+81+".."+(t_(n)-t_(n-1))-t_(n)`
or `t_(n)=1+3+9+27+81+".." " upto n terms "`
`(1*(3^(n)-1))/((3-1))=(1)/(2)(3^(n)-1)`
` therefore T_(n)=sumt_(n)=(1)/(2)(sum3^(n)-sum1)`
`=(1)/(2){(3+3^(2)+3^(3)+"..."+3^(n))-n}`
`=(1)/(2){(3(3^(n)-1))/((3-1))-n}`
`=(3)/(4)(3^(n)-1)-(1)/(2)n`
Discussion
503.

Find the sum of the series `1*n+2*(n-1)+3*(n-2)+4*(n-3)+....(n−1).2+n.1"` also, find the coefficient of `x^(n-1)` in th cxpansion of `(1+2x+3x^(2)+"...."nx^(n-1))^(2)`.

Answer» The rth term of the given series is
`T_(r)=r*(n-r+1)=(n+1)r-r^(2)`
`therefore` Sum of the series
`S_(n)=sum _(r=1)^(n)T_(r)=(n+1)sum _(r=1)^(n)r-sum _(r=1)^(n)r^(2)=(n+1)sumn-sumn^(2)`
`=(n+1)(n(n+1))/(2)-(n(n+1)(2n+1))/(6)`
`=(n(n+1))/(6)(3n+3-2n-1)=(n(n+1)(n+2))/(6)`
Now,
`(1+2x+3x^(2)+"..."+nx^(n-1))^(2)=(1+2x+3x^(2)+"..."+nx^(n-1))xx(1+2x+3x^(2)+"..."+nx^(n-1))`
`therefore " Coeffocient of "x^(n-1) " in "(1+2x+3x^(2)+"..."+nx^(n-1))^(2)`
`1*n+2*(n-1)+3*(n-2)+"..."+n*1`
`S_(n)=(n(n+1)(n+2))/(6)`
Discussion
504.

First term of a sequence is 1 and the `(n+1)th` term is obtained by adding `(n+1)` to the nth term for all natural numbers n, the 6th term of the sequence isA. 7B. 13C. 21D. 27

Answer» Correct Answer - C
Discussion
505.

Prove that: `2^(1/4).4^(1/8), 8^(1/16). 16^(1/32)......... oo=2`.A. 1B. 2C. `3//2`D. `5//2`

Answer» Correct Answer - B
We have,
`2^(1//4)*4^(1//8)*8^(1//16)*16^(1//32) . . .. . .oo`
`=2^(1//4+2//8+3//16+4//32_. .. . . . . . .oo`

`=2^(2^(1/2+)2^(2/3+)2^(3/4+)2^(4/5+) . . . . . . . .`
`=2^(2^(1/2){1+(2)/(2)+(3)/(2^(2))+(4)/(2^(3))}`
`2^(2^(1/2){(1)/(1-(1)/(2))+(1xx1//2)/(1-(1)/(2))^(2)}_(=2+(1)/(4)(2+2)_(=2^(1)=2)))`
Discussion
506.

Find the nth term and sum to n tems of the following series: 1+5+12+22+…………..

Answer» The sequence of differences between successuve terms is `4,7,10,13,"….",`.Clearly, it is an AP is in common difference 3. S0, let the nth rerm of the given series be `T_(n)` and sum of n temrs be `S_(n)`.
Then, `S_(n)=1+5+12+22+35+"..."+T_(n-1)+T_(n) "...(i)"`
`S_(n)=1+5+12+22+"..."+T_(n-1)+T_(n) "...(ii)"`
Subtracting Eq. (ii) from Eq. (i), we get
`0=1+4+7+10+13+"..."+(T_(n)-T_(n-1))-T_(n)`
`implies T_(n)=1+4+7+10+13+"..." " terms "`
`(n)/(2){2*1+(n-1)3}=(1)/(2)(3n^(2)-n)`
Hence, `T_(n)=(3)/(2)n^(2)-(1)/(2)n`
`therefore` Sum of n terms `S_(n)=sum T_(n)=(3)/(2)sumn^(2)-(1)/(2)sumn`
`=(3)/(2)((n(n+1)(2n+1))/(6))-(1)/(2)(n(n+1))/(2)`
`=(n(n+1))/(4)(2n+1-1)`
`=(1)/(2)n^(2)(n+1)=(1)/(2)(n^(3)+n^(2))`
Discussion
507.

Find the nth term of the series `=1+4+10+20+35+".."`.

Answer» The sequence of first consctive differences is `3,6,10,15,"…"` and second consecutive differences is `3,4,5,"….",`. Clearly, it is an AP with comon difference 1. So, let the nth term and sum of the seies upto n terms of the series be `T_(n)` and `S_(n)`, respectively.
Then, `S_(n)=1+4+10+20+35+".."+T_(n-1)+T_(n)"....(i)"`
`S_(n)=1+4+10+20+".."+T_(n-1)+T_(n)"....(ii)"`
Subtracting Eq. (ii) from Eq.(i), we get
`0=1+3+6+10+15+".."+(T_(n)-T_(n))-T_(n)`
`implies T_(n)=1+3+6+10+15+".." " upto n terms "`
or ` T_(n)=1+3+6+10+15+".."+t_(n-1)+t_(n)"....(iii)"`
` T_(n)=1+3+6+10+".."+t_(n-1)+t_(n)"....(iv)"`
Now, subtracting Eq. (iv) from Eq. (iii), we get ` 0=1+2+3+4+5+".."+(t_(n)-t_(n-1))-t_(n)`
or `t_(n)=1+2+3+4+5+".." " upto n terms "`
`=sum n=(n(n+1))/(2)`
` therefore T_(n)=sumt_(n)=(1)/(2)(sumn^(2)+sumn)`
`(1)/(2)((n(n+1)(2n+1))/(2)+(n(n+1))/(2))`
`(1)/(2)*(n(n+1))/(6)(2n+1+3)=(1)/(6)n(n+1)(n+2)`
Discussion
508.

Prove that: `2^(1/4).4^(1/8), 8^(1/16). 16^(1/32)......... oo=2`.A. 1B. `(3)/(2)`C. 2D. `(5)/(2)`

Answer» Correct Answer - D
Discussion
509.

`1+3+7+15+31+...+ ` to n termsA. `2^(n+1)-n`B. `2^(n+1)-n-2`C. `2^(n)-n-2`D. None of these

Answer» Correct Answer - B
Discussion
510.

Find the sum of the series `1+3+7+15+31+......` n terms.

Answer» The sequence of differences berween successive terms is `2,4,8,16,"…."`. Clearly, it is a GP with common ratio 2. So, let the nth term and sum of the series upto n terms of the series be `T_(n)` and `S_(n)`, respectively. Then,
`S_(n)=1+3+7+15+31+"..."T_(n-1)+T_(n) "....(i)"`
`S_(n)=1+3+7+15+"..."T_(n-1)+T_(n) "....(ii)"`
Subtracting Eq.(ii) from Eq.(i), we get
`0=1+2+4+8+16+"..."(T_(n)-T_(n-1))-T_(n)`
` implies T_(n)=1+2+4+8+16+"..." " upto n terms "`
` =(1*(2^(n)-1))/(2-1)`
Hence, `T_(n)=(2^(n)-1)`
`therefore` Sum of n terms `S_(n)=sumT_(n)=sum (2^(n)-1)=sum2^(n)-sum 1`
`=(2+2^(2)+2^(3)+"...."+2^(n))-n`
`=(2*(2-1))/(2-1)-n=2^(n+1)-2-n`
Discussion
511.

If `a_(n)=1+(1)/(2)+(1)/(3)+(1)/(4)+(1)/(5)+ . . . .+(1)/(2^(n)-1)`, thenA. `a_(100)lt100`B. `a_(100)gt100`C. `a_(200)lt100`D. none of these

Answer» Correct Answer - A
Discussion
512.

If a,b, and c are in G.P then a+b,2b and b+ c are inA. A.P.B. G.P.C. H.P.D. none of these

Answer» Correct Answer - C
Discussion
513.

If a,b,c are in A.P., a,x,b are in G.P. and b,y,c are in G.P. then `a^(2),b^(2),y^(2)` are inA. H.P.B. G.P.C. A.P.D. none of these

Answer» Correct Answer - C
Discussion
514.

`1/a+1/c+1/(a-b)+1/(c-b)=0` and `b!=a+c,` then `a,b,c ` are inA. H.P.B. G.P.C. A.P.D. none of these

Answer» Correct Answer - B
Discussion
515.

If three positive real numbers a,b,c, `(cgta)` are in H.P., then `log(a+c)+log(a-2b+c)` is equal toA. 2 log (c-b)B. 2 log (a+c)C. 2 log (c-a)D. log a+log b+log c

Answer» Correct Answer - B
Discussion
516.

If `log(x+z)+log(x-2y+z)=2log(x-z)," then "x,y,z` are inA. H.P.B. G.P.C. A.P.D. none of these

Answer» Correct Answer - A
Discussion
517.

If `x,y,z` are in H.P then the value of expression `log(x+z)+log(x-2y+z)=`A. log (x-z)B. 2log(x-z)C. 3log(x-z)D. 4log(x-z)

Answer» Correct Answer - B
Discussion
518.

If a, b, c are in G.P, then `log_a x, log_b x, log_c x` are inA. A.P.B. G.P.C. H.P.D. none of these

Answer» Correct Answer - C
Discussion