Question

Let a_n = (log 3)ⁿ$\sum_{r = 1}^{n}\frac{r^{2}}{r!(n - r)!}$, n ĪN. Then the sum of the series a₁ + a₂ + a₃ + a₄ + …. to  is –

• A. (1 + log 3) (9 log 3)
• B. (1 + log 3)
• C. (1 – log 3) (9 log 3)
• D. (1 + log 3) (log 3)

Answer 1

Answer: (A)

(1 + log 3) (9 log 3)

Answer 2

We have,

a_n = (log 3)ⁿ $\sum_{r = 1}^{n}\frac{r^{2}}{r!(n - r)!}$

= $\frac { ( \log 3 ) ^ { n } } { n ! }$ $\sum_{r = 1}^{n}{\frac{n!}{(n - r)!r!}r^{2}}$

= $\frac { ( \log 3 ) ^ { n } } { n ! }$ $\sum_{r = 1}^{n}{nC_{r}.r^{2}}$

= $\frac{(\log 3)^{n}}{n!}$ [n(n – 1) 2^n–2 + n · 2^n–1]

[Q $\sum_{r = 1}^{n}r^{2}$· ⁿC_r = n(n–1) 2^n–2 + n · 2^n–1]

= $\frac{(\log 3)^{n}}{n!}$ [n (n – 1) + 2n] 2^n–2

= $\frac{1}{(n - 2)!}$ (log 3)ⁿ · 2^n–2 + $\frac{2}{(n - 1)!}$ (log 3)ⁿ · 2^n–2

= (log 3)² $\frac{(2\log 3)^{n - 2}}{(n - 2)!}$+ (log 3) $\frac{(2\log 3)^{n - 1}}{(n - 1)!}$

\ Required sum

= $\sum_{n = 1}^{\infty}a_{n}$

= (log 3)² $\sum_{n = 2}^{\infty}\frac{(\log 9)^{n - 2}}{(n - 2)!}$+ (log 3) $\sum_{n = 1}^{\infty}\frac{(\log 9)^{n - 1}}{(n - 1)!}$

= (log 3)² e^{log 9} + (log 3) e^{log 9}

$\left\lbrack \because\sum_{n = 1}^{\infty}{\frac{x^{n - 1}}{(n - 1)!} = \sum_{n = 2}^{\infty}{\frac{x^{n - 2}}{(n - 2)!} = e^{x}}} \right\rbrack$

= (1 + log 3) (9 log 3)

Hence (1) is correct answe