`(2)/(2!)+(2+4)/(3!)+(2+4+6)/(4!)+….infty` is equal toA. `e-2`B. `e-1`C. eD. `e^(-1)`

`(2)/(2!)+(2+4)/(3!)+(2+4+6)/(4!)+….infty` is equal toA. `e-2`B. `e-

1.	`(2)/(2!)+(2+4)/(3!)+(2+4+6)/(4!)+….infty` is equal toA. `e-2`B. `e-1`C. eD. `e^(-1)`
Answer» Answer: Let `T_(n)` be the `n^(th)` term of the given series then `T_(n)=(2+4+6+..+2n)/(n+1)!,n=1,2,3………` `rarr T_(n)=(n(n+1))/(n+1)!=(1)/(n-1)!,n=1,2…` `therefore "Sum of the series" =underset(n=1)overset(infty)Sigma T_(n)=underset(n=1)overset(infty)Sigma (1)/(n-1)!=e`

`e^{(x-1)-1/2(x-1)^2+((x-1)^3)/3-(x-1)^(4)/4+......}` is eqaul toA. `log(x-1)`B. `logx`C. xD. none of these
`1+(log_e n)^2 /(2!) + (log_e n )^4 / (4!)+...=`A. nB. `(1)/(n)`C. `(n+n^(-1))/(2)`D. `(e^(n)+e^(-n))/(2)`
If n is not a multiple of 3, then the coefficient of `x^n` in the expansion of `log_e (1+x+x^2)`is : (A) `1/n` (B) `2/n` (C) `-1/n` (D) `-2/n`A. `-(2)/(n)`B. `(1)/(n)`C. `(2)/(n)`D. none of these
The sum of the series `1/2x^2+2/3x^3+3/4x^4+4/5x^5+`... is :A. `(x)/(1+x)+log(1+x)`B. `(x)/(1-x)+log(1-x)`C. `-(x)/(1+x)+log(1+x)`D. none of these
`(1)/(n!)+(1)/(2!(n-2)!)+(1)/(4!(n-4)!)`+….to n terms is equal toA. `(2n-1)/(2!)`B. `(2^(n))/(n+1)!`C. `(2^(n))/(n!)`D. `(2^(n)-2)/(n-1)!`
The coefficient of `x^(6)` in the expansion of `log{(1+x)^(1+x)(1-x)^(1-x)`}` isA. `(1)/(15)`B. `(1)/(30)`C. `(1)/(10)`D. `(1)/(45)`
The value of `9+(16)/(2!)+(27)/(3!)+(42)/(4!)+……infty` isA. `9e-6`B. `11e-6`C. `13e-6`D. `12e-6`
Sum of n terms of the series `1/(1.2.3.4.)+1/(2.3.4.5) +1/(3.4.5.6)+....`A. `1/6log_(e)2-(1)/(24)`B. `5/2-log_(e)2`C. `3/2-log_(e)2`D. none of these
The sum of series ` 2[ 7^(-1)+3^(-1).7^(-3)+5^(-1).7^(-5)+...]` isA. `log_(e)(4/3)`B. `log_(e)(3/4)`C. `2log_(e)(3/4)`D. `2 loge(4/3)`
The value of `log_(e) e- log_(9) e + log_(27) e- log_(81) e+…infty` isA. `log_(2)3`B. `log_(3)2`C. `log_(10)e`D. `log_(e)2`