The sum of the series 1+1(1+2)+1(1+2+3)+…∞ is :

Given: 1+1(1+2)+1(1+2+3)+…∞ ∴tn=1n(n+1)2=2n(n+1) ⇒tn=2n−2n+1,t1=21−22,t2=22−23t3=23−24tn−1=2n−1−2n∴ Σtn=t1+t2+….+tn=2−2n+1⇒S∞=limn→∞ tn=limn→∞ 2−2n+1=2−0=2

The sum of the series 1+1(1+2)+1(1+2+3)+…∞ is :

Given: 1+1(1+2)+1(1+2+3)+…∞ ∴tn=1n(n+1)2=2n(n+1) ⇒tn=2n−2n+1,t1=21−22,t2=22−23t3=23−24tn−1=2n−1−2n∴ Σtn=t1+t2+….+tn=2−2n+1⇒S∞=limn→∞ tn=limn→∞ 2−2n+1=2−0=2

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The sum of the series $1 + \frac{1}{(1 + 2)} + \frac{1}{(1 + 2 + 3)} + \dots \infty$ is :

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$\frac{1}{2}$

$\frac{3}{2}$

answer is C.

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Detailed Solution

Given: $1 + \frac{1}{(1 + 2)} + \frac{1}{(1 + 2 + 3)} + \dots \infty ∴ t_{n} = \frac{1}{\frac{n (n + 1)}{2}} = \frac{2}{n (n + 1)} \begin{array}{l} \Rightarrow t_{n} = \frac{2}{n} - \frac{2}{n + 1}, t_{1} = \frac{2}{1} - \frac{2}{2}, t_{2} = \frac{2}{2} - \frac{2}{3} \\ t_{3} = \frac{2}{3} - \frac{2}{4} \\ t_{n - 1} = \frac{2}{n - 1} - \frac{2}{n} \\ ∴ {Σt}_{n} = t_{1} + t_{2} + \dots . + t_{n} = 2 - \frac{2}{n + 1} \\ \Rightarrow S_{\infty} = lim_{n \to \infty} t_{n} = lim_{n \to \infty} (2 - \frac{2}{n + 1}) \\ = 2 - 0 = 2 \end{array}$