The sum of the series (2)^2 + 2(4)^2 + 3(6)^2 | vedclass

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

Question

(C) The given series is $S = 1(2)^2 + 2(4)^2 + 3(6)^2 + ...$ up to $10$ terms.The $n$-th term of the series is $T_n = n(2n)^2 = n(4n^2) = 4n^3$.To fin

Vedclass · Accepted Answer

$12100$

Vedclass · Answer

(C) The given series is $S = 1(2)^2 + 2(4)^2 + 3(6)^2 + ...$ up to $10$ terms.The $n$-th term of the series is $T_n = n(2n)^2 = n(4n^2) = 4n^3$.To find the sum of $10$ terms,we calculate $\sum_{n=1}^{10} 4n^3$.$S = 4 \sum_{n=1}^{10} n^3$.Using the formula for the sum of cubes of first $n$ natural numbers,$\sum_{n=1}^{n} n^3 = [\frac{n(n+1)}{2}]^2$.For $n = 10$,$\sum_{n=1}^{10} n^3 = [\frac{10 	imes 11}{2}]^2 = (55)^2 = 3025$.Therefore,$S = 4 	imes 3025 = 12100$.

The sum of the series $(2)^2 + 2(4)^2 + 3(6)^2 + ...$ up to $10$ terms is:

Similar Questions

In a $G.P.$,the first term is $7$,the last term is $448$,and the sum is $889$. Find the common ratio.

The fifth term of the $H.P.$,$2, 2\frac{1}{2}, 3\frac{1}{3}, \dots$ will be

If the sum of first $n$ terms of an $A.P.$ is $cn(n - 1)$,where $c \neq 0$,then the sum of the squares of these terms is

If the $7^{th}$ term of a harmonic progression is $8$ and the $8^{th}$ term is $7$,then its $15^{th}$ term is

The sum of the series $3 + 33 + 333 + \dots + n$ terms is

Vedclass Test Series

Exam Paper Generator

Online Exam Module