The sum of the series,frac{1}{2 cdot 3} cdot | vedclass

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(B) The $n$-th term of the series is given by $T_n = \frac{n}{(n+1)(n+2)} \cdot 2^n$. Using partial fractions,we can write $\frac{n}{(n+1)(n+2)} = \fr

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$\frac{2^{n+1}}{n+2} - 1$

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(B) The $n$-th term of the series is given by $T_n = \frac{n}{(n+1)(n+2)} \cdot 2^n$. Using partial fractions,we can write $\frac{n}{(n+1)(n+2)} = \frac{2}{n+2} - \frac{1}{n+1}$. Thus,$T_n = \left( \frac{2}{n+2} - \frac{1}{n+1} \right) 2^n = \frac{2^{n+1}}{n+2} - \frac{2^n}{n+1}$. Let $f(n) = \frac{2^n}{n+1}$. Then $T_n = f(n+1) - f(n)$. The sum $S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n (f(k+1) - f(k)) = f(n+1) - f(1)$. Since $f(n+1) = \frac{2^{n+1}}{n+2}$ and $f(1) = \frac{2^1}{1+1} = 1$,we have $S_n = \frac{2^{n+1}}{n+2} - 1$.

The sum of the series,$\frac{1}{2 \cdot 3} \cdot 2 + \frac{2}{3 \cdot 4} \cdot 2^{2} + \frac{3}{4 \cdot 5} \cdot 2^{3} + \ldots$ up to $n$ terms is

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