If S_n = (1 · 2)/(3!) + (2 · 2^2)/(4!) + (3 · 2^3)/(5!) + upto n terms then the sum of the infinite terms is:

Correct answer: B. (2)/(3) The r^ th term is T_r = (r · 2^r)/((r+2)!) . We can rewrite the numerator: r · 2^r = (r+2-2)2^r = (r+2)2^r - 2^ r+1 . So, T_r = ((r+2)2^r)/((r+2)!) - 2^ r+1 (r+2)! = (2^r)/((r+1)!) - 2^ r+1 (r+2)! . Now, calculate the sum S_n = ∑_ r=1 ^n T_r : T_1 = (2^1)/(2!) - (2^2)/(3!)T_2 = (2^2)/(3!) - (2^3)/(4!)T_n = (2^n)/((n+1)!) - 2^ n+1 (n+2)! Summing these terms (telescoping series): S_n = (2^1)/(2!) - 2^ n+1 (n+2)! = 1 - 2^ n+1 (n+2)! . As n → ∞ , the term 2^ n+1 (n+2)! → 0 . Thus, S_∞ = 1 - 0 = 1 .

Black Advanced Problems in Mathematics

Exam: Black Advanced Problems in Mathematics
Subject: Mathematics
Chapter: Sequences and Series

Black Advanced Problems in Mathematics · Mathematics · Sequences and Series

If S_n = (1 · 2)/(3!) + (2 · 2^2)/(4!) + (3 · 2^3)/(5!) + upto n...

S_n = \frac{1 \cdot 2}{3!} + \frac{2 \cdot 2^2}{4!} + \frac{3 \cdot 2^3}{5!} + \dots

upto

n

terms then the sum of the infinite terms is:

Options

A $1$
B $\frac{2}{3}$
C $e$
D $\frac{\pi}{4}$

Correct answer

B. $\frac{2}{3}$

Step-by-step solution

The

r^{\text{th}}

term is

T_r = \frac{r \cdot 2^r}{(r+2)!}

. We can rewrite the numerator:

r \cdot 2^r = (r+2-2)2^r = (r+2)2^r - 2^{r+1}

. So,

T_r = \frac{(r+2)2^r}{(r+2)!} - \frac{2^{r+1}}{(r+2)!} = \frac{2^r}{(r+1)!} - \frac{2^{r+1}}{(r+2)!}

. Now, calculate the sum

S_n = \sum_{r=1}^n T_r

T_1 = \frac{2^1}{2!} - \frac{2^2}{3!}T_2 = \frac{2^2}{3!} - \frac{2^3}{4!}T_n = \frac{2^n}{(n+1)!} - \frac{2^{n+1}}{(n+2)!}

Summing these terms (telescoping series):

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

. As

n \to \infty

, the term

\frac{2^{n+1}}{(n+2)!} \to 0

. Thus,

S_\infty = 1 - 0 = 1

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