Sum of the series 2² + 4² + 6² + ....+ 20² is

Question

Question

This question was previously asked in

UPSC ESE 2021 Official Paper 1

Answer 1

Concept:

We know

Sum of square of n integers is given by

12 + 22 + 32 + ..... + n2 = \(\frac{n(n+1)(2n+1)}{6}\)

Calculation:

Here we have to find the sum of the series 22 + 42 + 62 + ....+ 202

The given series can be re-written as: 22 ×(12 + 22 + 32 + ..... + 102)

As we know that, 12 + 2 2 + 32 + ..... + n2 = \(\frac{n(n+1)(2n+1)}{6}\)

Here, n = 10

⇒ 12 + 22 + 32 + ..... + 102 = \(\frac{10\; × \;11 \;× \;21}{6} = 385\)

So, Sum of 22 × (12 + 22 + 32 + ..... + 102) = 22 × 385 = 1540