The sum displaystyle∑n=1∞ (2 n2+3 n+4/(2 n) !) is equal to:

Question

The sum displaystyle∑n=1∞ (2 n2+3 n+4/(2 n) !) is equal to: (A) (11 e /2)+(7/2 e )-4 (B) (11 e /2)+(7/2 e ) (C) (13 e /4)+(5/4 e )-4 (D) (13 e /4)+(5/4 e )

Tardigrade Admin · Accepted Answer

displaystyle∑n=1∞ (2 n2+3 n+4/(2 n) !) (1/2) displaystyle∑n=1∞ (2 n(2 n-1)+8 n+8/(2 n) !) (1/2) displaystyle∑n=1∞ (1/(2 n-2) !)+2 displaystyle∑n=1∞ (1/(2 n-1) !)+4 displaystyle∑n=1∞ (1/(2 n) !) e =1+1+(1/2 !)+(1/3 !)+(1/4 !)+ ldots . e -1=1-1+(1/2 !)-(1/3 !)+(1/4 !)+ ldots . ( e +(1/ e ))=2(1+(1/2 !)+(1/4 !)+ ldots ldots .) e -(1/ e )=(1+(1/3 !)+(1/5 !)+ ldots . .) Now (1/2)( displaystyle∑n=1∞ (1/(2 n-2) !))+2 displaystyle∑n=1∞ (1/(2 n-1) !)+4 displaystyle∑n=1∞ (1/(2 n) !) =(1/2)[(e+(1/e)/2)]+2[(e-(1/e)/2)]+4((e+(1/e)-2/2)) =((e+(1/e))/4)+e-(1/e)+2 e+(2/e)-4 =(13/4) e+(5/4 e)-4

Q. The sum $n = 1 \sum \infty \frac{2 n ^{2} + 3 n + 4}{( 2 n )!}$ is equal to:

$\frac{11 e}{2} + \frac{7}{2 e} - 4$

$\frac{11 e}{2} + \frac{7}{2 e}$

$\frac{13 e}{4} + \frac{5}{4 e} - 4$

$\frac{13 e}{4} + \frac{5}{4 e}$

Q. The sum n=1∑∞​(2n)!2n2+3n+4​ is equal to:

211e​+2e7​−4

211e​+2e7​

413e​+4e5​−4

413e​+4e5​

Q. The sum $n = 1 \sum \infty \frac{2 n ^{2} + 3 n + 4}{( 2 n )!}$ is equal to:

$\frac{11 e}{2} + \frac{7}{2 e} - 4$

$\frac{11 e}{2} + \frac{7}{2 e}$

$\frac{13 e}{4} + \frac{5}{4 e} - 4$

$\frac{13 e}{4} + \frac{5}{4 e}$