Find the sum of the series 2² + 4² + 6² + .... + (2n)²

Question

Question

Answer 1

Concept:

Sum of the squares of first n natural numbers is given by:

\(\sum_{k=1}^{n}k^2=1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}\)

Calculation:

Let T_n be the nth term of the series. Then,

T_n = (2n)²= 4n²

∴ 22 + 42 + 62 + .... + (2n)² = \(\sum_{k=1}^{n}T_{k}\)

\(\Rightarrow \sum_{k=1}^{n}4k^2\)

\(\Rightarrow 4\sum_{k=1}^{n}k^2\)

\(\Rightarrow 4\left \{ \frac{n(n+1)(2n+1)}{6} \right \}\)

\(\Rightarrow \frac{2}{3}\: n(n+1)(2n+1)\)

Hence, the sum of the series 22 + 42 + 62 + .... + (2n)2 is

\(\frac{2}{3}\: n(n+1)(2n+1)\).

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