Find the sum to n terms of the series: frac{3 | vedclass

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(A) The $n^{th}$ term is $t_n = \frac{n+2}{n(n+1)} \cdot \left( \frac{1}{2} \right)^n$. We can write $\frac{n+2}{n(n+1)} = \frac{2(n+1) - n}{n(n+1)} =

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

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(A) The $n^{th}$ term is $t_n = \frac{n+2}{n(n+1)} \cdot \left( \frac{1}{2} \right)^n$. We can write $\frac{n+2}{n(n+1)} = \frac{2(n+1) - n}{n(n+1)} = \frac{2}{n} - \frac{1}{n+1}$. Thus,$t_n = \left( \frac{2}{n} - \frac{1}{n+1} \right) \left( \frac{1}{2} \right)^n = \frac{1}{n} \left( \frac{1}{2} \right)^{n-1} - \frac{1}{n+1} \left( \frac{1}{2} \right)^n$. Let $f(n) = \frac{1}{n} \left( \frac{1}{2} \right)^{n-1}$. Then $t_n = f(n) - f(n+1)$. The sum $S_n = \sum_{k=1}^n t_k = (f(1) - f(2)) + (f(2) - f(3)) + \dots + (f(n) - f(n+1)) = f(1) - f(n+1)$. $f(1) = \frac{1}{1} \left( \frac{1}{2} \right)^0 = 1$. $f(n+1) = \frac{1}{n+1} \left( \frac{1}{2} \right)^n$. Therefore,$S_n = 1 - \frac{1}{(n+1) 2^n}$.

Find the sum to $n$ terms of the series: $\frac{3}{1 \cdot 2} \cdot \frac{1}{2} + \frac{4}{2 \cdot 3} \cdot \left( \frac{1}{2} \right)^2 + \frac{5}{3 \cdot 4} \cdot \left( \frac{1}{2} \right)^3 + \dots$

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