If sum_{n = 1}^5 frac{1}{n(n + 1)(n + 2)(n + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The general term of the given expression can be written using the method of differences as:$T_n = \frac{1}{3} \left[ \frac{1}{n(n + 1)(n + 2)} - \

Vedclass · Accepted Answer

$\frac{55}{336}$

Vedclass · Answer

(C) The general term of the given expression can be written using the method of differences as:$T_n = \frac{1}{3} \left[ \frac{1}{n(n + 1)(n + 2)} - \frac{1}{(n + 1)(n + 2)(n + 3)} \right]$Taking the summation from $n = 1$ to $5$ on both sides,we get a telescoping series:$\sum_{n = 1}^5 T_n = \frac{1}{3} \left[ \left( \frac{1}{1 \cdot 2 \cdot 3} - \frac{1}{2 \cdot 3 \cdot 4} \right) + \dots + \left( \frac{1}{5 \cdot 6 \cdot 7} - \frac{1}{6 \cdot 7 \cdot 8} \right) \right]$This simplifies to:$\sum_{n = 1}^5 T_n = \frac{1}{3} \left[ \frac{1}{6} - \frac{1}{6 \cdot 7 \cdot 8} \right] = \frac{k}{3}$$\Rightarrow \frac{1}{3} \left[ \frac{1}{6} - \frac{1}{336} \right] = \frac{k}{3}$$\Rightarrow \frac{1}{6} - \frac{1}{336} = k$$\Rightarrow k = \frac{56 - 1}{336} = \frac{55}{336}$

If $\sum_{n = 1}^5 \frac{1}{n(n + 1)(n + 2)(n + 3)} = \frac{k}{3}$,then $k$ is equal to

Similar Questions

Find the common difference of an $A.P.$ whose first term is $100$ and the sum of whose first six terms is five times the sum of the next six terms.

The sum of $n$ terms of the series whose $n^{th}$ term is $n(n + 1)$ is equal to

The ratio of the sum of $m$ and $n$ terms of an $A.P.$ is $m^2 : n^2$. Then the ratio of the $m^{th}$ and $n^{th}$ term will be:

If the $n^{th}$ term of a series is $3 + n(n - 1)$,then the sum of $n$ terms of the series is

If $\log 2, \log (2^n - 1)$ and $\log (2^n + 3)$ are in $A.P.$,then $n =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module