Ifn sum _n = 1^5mspace2mu(1 / n(n + 1)(n + 2)(n + 3)) = (k / 3),n then k is equal to

(c) General term of given expression can be written asnn T_ r=(1 / 3)[frac1 n( n+1)( n+2)-frac1( n+1)( n+2)( n+3)]nnon taking summation both the side, we getnn n sum _ r=1^5 T_ r=(1 / 3)[(1 / 6)-(1 / 6.7 .8)]=frac k3 n => (1 / 3) x (1 / 6)(1-(1 / 56))=frac k3 => (1 / 3) x (1 / 6) x (55 / 56)=frac k3 n => k=(55 / 336)n n

If \sum_{n = 1}^{5}\mspace{2mu}\frac{1}{n(n + 1)(n + ... | Sequence & Series (Mathematics)

\n\mathrm{T}_{\mathrm{r}}=\frac{1}{3}\left[\frac{1}{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}-\frac{1}{(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)}\right]\n

\non taking summation both the side, we get\n

\n\begin{aligned}\n& \sum_{\mathrm{r}=1}^5 \mathrm{~T}_{\mathrm{r}}=\frac{1}{3}\left[\frac{1}{6}-\frac{1}{6.7 .8}\right]=\frac{\mathrm{k}}{3} \\\n& \Rightarrow \frac{1}{3} \times \frac{1}{6}\left(1-\frac{1}{56}\right)=\frac{\mathrm{k}}{3} \Rightarrow \frac{1}{3} \times \frac{1}{6} \times \frac{55}{56}=\frac{\mathrm{k}}{3} \\\n& \Rightarrow \mathrm{k}=\frac{55}{336}\n\end{aligned}\n

Explanation

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