The sum frac{3}{1^2} + frac{5}{1^2 + 2^2} + f | vedclass

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

Question

(C) The given series is $\sum_{n=1}^{11} T_n$,where the $n^{th}$ term $T_n$ is given by:$T_n = \frac{2n+1}{1^2 + 2^2 + \dots + n^2}$Using the formula

Vedclass · Accepted Answer

$\frac{11}{2}$

Vedclass · Answer

(C) The given series is $\sum_{n=1}^{11} T_n$,where the $n^{th}$ term $T_n$ is given by:$T_n = \frac{2n+1}{1^2 + 2^2 + \dots + n^2}$Using the formula for the sum of squares of the first $n$ natural numbers,$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$,we get:$T_n = \frac{2n+1}{\frac{n(n+1)(2n+1)}{6}} = \frac{6}{n(n+1)}$Using partial fractions,we can write:$T_n = 6 \left( \frac{1}{n} - \frac{1}{n+1} ight)$Now,the sum of $n$ terms $S_n$ is:$S_n = \sum_{k=1}^n 6 \left( \frac{1}{k} - \frac{1}{k+1} ight) = 6 \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n+1}) ight]$$S_n = 6 \left( 1 - \frac{1}{n+1} ight) = \frac{6n}{n+1}$For $n=11$ terms:$S_{11} = \frac{6 	imes 11}{11+1} = \frac{66}{12} = \frac{11}{2}$

The sum $\frac{3}{1^2} + \frac{5}{1^2 + 2^2} + \frac{7}{1^2 + 2^2 + 3^2} + \dots$ up to $11$ terms is

Similar Questions

If $n$ is odd or even,then the sum of $n$ terms of the series $1 - 2 + 3 - 4 + 5 - 6 + \dots$ will be

The sum of the series $\frac{1}{{1 + {1^2} + {1^4}}} + \frac{2}{{1 + {2^2} + {2^4}}} + \frac{3}{{1 + {3^2} + {3^4}}} + \dots$ to $n$ terms is

If a clock strikes the appropriate number of times at each hour,how many times will it strike in a day?

Determine the sum of the first $35$ terms of an $A.P.$ if $t_{2}=2$ and $t_{7}=22$.

If $x = \sum_{n = 0}^\infty a^n$,$y = \sum_{n = 0}^\infty b^n$,and $z = \sum_{n = 0}^\infty (ab)^n$,where $a, b < 1$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module