frac{1 cdot 2}{1!} + frac{2 cdot 3}{2!} + fra | vedclass

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(B) The general term of the series is $T_n = \frac{n(n+1)}{n!} = \frac{n+1}{(n-1)!}$.We can rewrite the numerator as $(n-1) + 2$,so $T_n = \frac{(n-1)

Vedclass · Accepted Answer

$3e$

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(B) The general term of the series is $T_n = \frac{n(n+1)}{n!} = \frac{n+1}{(n-1)!}$.We can rewrite the numerator as $(n-1) + 2$,so $T_n = \frac{(n-1) + 2}{(n-1)!} = \frac{n-1}{(n-1)!} + \frac{2}{(n-1)!} = \frac{1}{(n-2)!} + \frac{2}{(n-1)!}$.Summing the series $S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \frac{1}{(n-2)!} + \sum_{n=1}^{\infty} \frac{2}{(n-1)!}$.Note that $\frac{1}{(-1)!} = 0$,so the first sum starts effectively from $n=2$ as $\sum_{n=2}^{\infty} \frac{1}{(n-2)!} = \sum_{k=0}^{\infty} \frac{1}{k!} = e$.The second sum is $2 \sum_{n=1}^{\infty} \frac{1}{(n-1)!} = 2 \sum_{k=0}^{\infty} \frac{1}{k!} = 2e$.Thus,$S = e + 2e = 3e$.

$\frac{1 \cdot 2}{1!} + \frac{2 \cdot 3}{2!} + \frac{3 \cdot 4}{3!} + \frac{4 \cdot 5}{4!} + \dots \infty = $

Similar Questions

The sum of the series $\sum_{n=1}^{\infty} \frac{n^{2}+6 n+10}{(2 n+1) !}$ is equal to :

The coefficient of $x^n$ in $\frac{1-2x}{e^x}$ is:

$\frac{x^2 - y^2}{1!} + \frac{x^4 - y^4}{2!} + \frac{x^6 - y^6}{3!} + \dots \infty = $

Find the sum of the series $\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots \infty$.

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

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