If the sum of the first 10 terms of the serie | vedclass

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

Question

(B) The $r$-th term of the series is given by $T_r = \frac{4r}{1+4r^4}$.Using the identity $1+4r^4 = (1+2r^2-2r)(1+2r^2+2r)$,we can write:$T_r = \frac

Vedclass · Accepted Answer

$441$

Vedclass · Answer

(B) The $r$-th term of the series is given by $T_r = \frac{4r}{1+4r^4}$.Using the identity $1+4r^4 = (1+2r^2-2r)(1+2r^2+2r)$,we can write:$T_r = \frac{(2r^2+2r+1)-(2r^2-2r+1)}{(2r^2+2r+1)(2r^2-2r+1)}$$T_r = \frac{1}{2r^2-2r+1} - \frac{1}{2r^2+2r+1}$.For $r=1, T_1 = \frac{1}{1} - \frac{1}{5}$.For $r=2, T_2 = \frac{1}{5} - \frac{1}{13}$.Continuing this up to $r=10$,$T_{10} = \frac{1}{2(10)^2-2(10)+1} - \frac{1}{2(10)^2+2(10)+1} = \frac{1}{181} - \frac{1}{221}$.The sum $S_{10} = \sum_{r=1}^{10} T_r = 1 - \frac{1}{221} = \frac{220}{221}$.Given $S_{10} = \frac{m}{n}$,we have $m=220$ and $n=221$.Since $\operatorname{gcd}(220, 221) = 1$,$m+n = 220+221 = 441$.

If the sum of the first $10$ terms of the series $\frac{4(1)}{1+4(1)^4}+\frac{4(2)}{1+4(2)^4}+\frac{4(3)}{1+4(3)^4}+\ldots$ is $\frac{m}{n}$,where $\operatorname{gcd}(m, n)=1$,then $m+n$ is equal to . . . . . . .

Similar Questions

If ${t_n} = \frac{1}{4}(n + 2)(n + 3)$ for $n = 1, 2, 3, \dots$,then $\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} + \dots + \frac{1}{t_{2003}} = $

For any $n \in N$,$\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \ldots + \frac{1}{(3n-1)(3n+2)} = $

The value of $\mathop {\lim }\limits_{n \to \infty } \left( \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2n - 1)(2n + 1)} \right)$ is equal to

The sum of the first $n$ terms of the series $\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \ldots$ is

The sum $1(1!) + 2(2!) + 3(3!) + \dots + n(n!)$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module