Question

Find the sum of the series $2^2+4^2+6^2+……+(2n) ^2$

Answer 1

Hint: To find the sum of the given series , one can use the formula to find the sum squares of first n natural numbers is $\dfrac{n(n+1)(2n+1)}{6}$

Complete step-by-step answer:
Given , sum of the series $2^2+4^2+6^2+……+(2n) ^2$ is a even sum where every number is divisible by 4
Thus , Even sum = $2^2+4^2+6^2+……+(2n) ^2$
 By , taking 4 common , we get
Even sum = $4 \times(1^2+2^2+3^2+……+(n) ^2$
Now , putting the formula for even sum , we get
Even sum = $\dfrac{4 \times n(n+1)(2n+1)}{6}$=$\dfrac{2n(n+1)(n+1)}{3}$
Thus , the required sum for the given series is $\dfrac{2n(n+1)(n+1)}{3}$

Note: Summation is the addition of a list, or sequence, of numbers. If the summation sequence contains an infinite number of terms, this is called a series. Sums and series are iterative operations that provide many useful and interesting results in the field of mathematics. Students first need to understand the meaning of series before solving the sum and then learn the way of adding such series.