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Sum to Infinity of r·2^r/(r+2)! Series | JEE

JEE Maths question with a full step-by-step solution.

Question

S_n=\frac{1\cdot2}{3!}+\frac{2\cdot2^2}{4!}+\frac{3\cdot2^3}{5!}+\cdots

up to

n

terms, then the sum to infinite terms is

\frac{4}{\pi}

\frac{3}{e}

\frac{\pi}{3}

1

correct

Solution

Step 1: General term

t_r=\dfrac{r\cdot 2^r}{(r+2)!}

. Write

r=(r+2)-2

t_r=\frac{(r+2)2^r-2\cdot 2^r}{(r+2)!}=\frac{2^r}{(r+1)!}-\frac{2^{r+1}}{(r+2)!}.

Step 2: Sum from

r=1

n

; it telescopes:

S_n=\frac{2^1}{2!}-\frac{2^{n+1}}{(n+2)!}=1-\frac{2^{n+1}}{(n+2)!}.

Step 3: As

n\to\infty

the second term

\to 0

, so

S_\infty=1.

Correct answer: (4)

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