If sum_{k=1}^{10} frac{k}{k^{4}+k^{2}+1}=frac | vedclass

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

Question

(A) We have the sum $\sum_{k=1}^{10} \frac{k}{k^{4}+k^{2}+1}$.Note that $k^{4}+k^{2}+1 = (k^{2}+1)^{2}-k^{2} = (k^{2}+k+1)(k^{2}-k+1)$.Thus,the genera

Vedclass · Accepted Answer

$166$

Vedclass · Answer

(A) We have the sum $\sum_{k=1}^{10} \frac{k}{k^{4}+k^{2}+1}$.Note that $k^{4}+k^{2}+1 = (k^{2}+1)^{2}-k^{2} = (k^{2}+k+1)(k^{2}-k+1)$.Thus,the general term is $\frac{k}{(k^{2}+k+1)(k^{2}-k+1)}$.Using partial fractions,we write $\frac{k}{(k^{2}+k+1)(k^{2}-k+1)} = \frac{1}{2} \left( \frac{1}{k^{2}-k+1} - \frac{1}{k^{2}+k+1} \right)$.Let $f(k) = \frac{1}{k^{2}-k+1}$. Then the sum is $\frac{1}{2} \sum_{k=1}^{10} (f(k) - f(k+1))$.This is a telescoping sum: $\frac{1}{2} (f(1) - f(11))$.$f(1) = \frac{1}{1^{2}-1+1} = 1$.$f(11) = \frac{1}{11^{2}-11+1} = \frac{1}{121-11+1} = \frac{1}{111}$.So,the sum is $\frac{1}{2} (1 - \frac{1}{111}) = \frac{1}{2} (\frac{110}{111}) = \frac{55}{111}$.Here $m=55$ and $n=111$,which are coprime.Therefore,$m+n = 55+111 = 166$.

If $\sum_{k=1}^{10} \frac{k}{k^{4}+k^{2}+1}=\frac{m}{n}$,where $m$ and $n$ are coprime,then $m+n$ is equal to.

Similar Questions

$1 + \sum\limits_{r = 0}^{22} {\left\{ {r\left( {r + 2} \right) + 1} \right\}} \cdot r! = k!$,then the number of divisors of $k$ is

The sum to $10$ terms of the series $\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+\ldots$ is:

The sum of $(n + 1)$ terms of $\frac{1}{1} + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \dots$ is

If ${a_k} = \frac{1}{{k(k + 1)}}$ for $k = 1, 2, 3, 4, ..., n$,then ${\left( {\sum\limits_{k = 1}^n {{a_k}} } \right)^2} = $

If $\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=m$ and $\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=n$,then the point $(m, n)$ lies on the line

Vedclass Test Series

Exam Paper Generator

Online Exam Module