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

If the sum of the first $2n$ terms of $2, 5, 8, \dots$ is equal to the sum of the first $n$ terms of $57, 59, 61, \dots$, then $n$ is equal to

The value of ${4^{1/3}} \cdot {4^{1/9}} \cdot {4^{1/27}} \cdots \infty$ is

The first and last terms of a $G.P.$ are $a$ and $l$ respectively; $r$ being its common ratio; then the number of terms in this $G.P.$ is

The value of $\sum_{n=0}^{\infty} \frac{(n+1)^2}{7^n}$ is -

Consider an infinite $G.P.$ with first term $a$ and common ratio $r$. If its sum is $4$ and the second term is $3/4$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module