If the sum of the first ten terms of the seri | vedclass

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

Question

(D) The series is $S = \left(\frac{8}{5}\right)^2 + \left(\frac{12}{5}\right)^2 + \left(\frac{16}{5}\right)^2 + \left(\frac{20}{5}\right)^2 + \dots$ u

Vedclass · Accepted Answer

$101$

Vedclass · Answer

(D) The series is $S = \left(\frac{8}{5}ight)^2 + \left(\frac{12}{5}ight)^2 + \left(\frac{16}{5}ight)^2 + \left(\frac{20}{5}ight)^2 + \dots$ up to $10$ terms.This can be written as $S = \frac{1}{25} \left( 8^2 + 12^2 + 16^2 + 20^2 + \dots + (4(n+1))^2 ight)$ for $n=1$ to $10$.$S = \frac{16}{25} \sum_{n=1}^{10} (n+1)^2$.Let $k = n+1$,then the sum is $\frac{16}{25} \sum_{k=2}^{11} k^2$.Using the formula $\sum_{k=1}^{N} k^2 = \frac{N(N+1)(2N+1)}{6}$,we have $\sum_{k=1}^{11} k^2 = \frac{11(12)(23)}{6} = 11 	imes 2 	imes 23 = 506$.Since the sum starts from $k=2$,we subtract $1^2 = 1$: $506 - 1 = 505$.So,$S = \frac{16}{25} 	imes 505 = \frac{16 	imes 101}{5} = \frac{16}{5} 	imes 101$.Given $S = \frac{16}{5}m$,we get $m = 101$.

If the sum of the first ten terms of the series ${\left( {1\frac{3}{5}} \right)^2} + {\left( {2\frac{2}{5}} \right)^2} + {\left( {3\frac{1}{5}} \right)^2} + {4^2} + \dots$ is $\frac{16}{5}m$,then $m$ is equal to:

Similar Questions

If $\sum_{k=1}^{n} a_k = 6n^3$, then $\sum_{k=1}^{6} \left(\frac{a_{k+1}-a_k}{36}\right)^2$ is equal to . . . . . . .

If $4^3+8^3+12^3+\ldots$ up to $n$ terms $= k n^2(n+1)^2$ (for all $n \in N$),then $k=$

The mean of the squares of the first $n$ natural numbers is

The number of positive integers $n$ in the set $\{1, 2, 3, \ldots, 100\}$ for which the number $\frac{1^2+2^2+3^2+\ldots+n^2}{1+2+3+\ldots+n}$ is an integer is

If $1 \cdot 3 \cdot 5 + 3 \cdot 5 \cdot 7 + 5 \cdot 7 \cdot 9 + \ldots \text{ to } n \text{ terms} = n(n+1) f(n) - 3n$,then $f(1) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module