1 + frac{1 + 2}{1!} + frac{1 + 2 + 3}{2!} + f | vedclass

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Question

(C) The $n^{th}$ term of the series is given by $T_n = \frac{1 + 2 + 3 + \dots + n}{(n-1)!}$.Using the sum of first $n$ natural numbers,$T_n = \frac{n

Vedclass · Accepted Answer

$\frac{7e}{2}$

Vedclass · Answer

(C) The $n^{th}$ term of the series is given by $T_n = \frac{1 + 2 + 3 + \dots + n}{(n-1)!}$.Using the sum of first $n$ natural numbers,$T_n = \frac{n(n+1)}{2(n-1)!}$.We can rewrite the numerator as $n(n+1) = (n-1+1)(n-1+2) = (n-1)^2 + 3(n-1) + 2$.Thus,$T_n = \frac{1}{2} \left[ \frac{(n-1)(n-2) + 3(n-1) + 2}{(n-1)!} \right] = \frac{1}{2} \left[ \frac{1}{(n-3)!} + \frac{3}{(n-2)!} + \frac{2}{(n-1)!} \right]$.Summing from $n=1$ to $\infty$,we get $S = \frac{1}{2} \left[ \sum_{n=3}^{\infty} \frac{1}{(n-3)!} + 3 \sum_{n=2}^{\infty} \frac{1}{(n-2)!} + 2 \sum_{n=1}^{\infty} \frac{1}{(n-1)!} \right]$.Since $\sum_{k=0}^{\infty} \frac{1}{k!} = e$,we have $S = \frac{1}{2} [e + 3e + 2e] = \frac{6e}{2} = 3e$. Wait,re-evaluating the series: $T_1 = 1/0! = 1$,$T_2 = 3/1! = 3$,$T_3 = 6/2! = 3$,$T_4 = 10/6 = 5/3$.The sum is $\sum_{n=1}^{\infty} \frac{n(n+1)}{2(n-1)!} = \frac{1}{2} \sum_{n=1}^{\infty} \frac{(n-1+1)(n-1+2)}{(n-1)!} = \frac{1}{2} \sum_{k=0}^{\infty} \frac{(k+1)(k+2)}{k!} = \frac{1}{2} \sum_{k=0}^{\infty} \frac{k^2+3k+2}{k!} = \frac{1}{2} [\sum \frac{k(k-1)+4k+2}{k!}] = \frac{1}{2} [e + 4e + 2e] = \frac{7e}{2}$.

$1 + \frac{1 + 2}{1!} + \frac{1 + 2 + 3}{2!} + \frac{1 + 2 + 3 + 4}{3!} + \dots \infty = $

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