Correct Answer - Option 1 : 506

__Concept:__

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

__Calculation:__

Here we have to find the sum of the series 1 + 4 + 9 +16 + 25 + 36 +......+ 121

The given series can be re-written as: 1^{2} + 2^{2} + 3^{2} + ..... + 11^{2}

As we know that, 12 + 22 + 32 + ..... + n2 = \(\frac{n(n+1)(2n+1)}{6}\)

Here, n = 11

⇒ 1^{2} + 2^{2} + 3^{2} + ..... + 11^{2} = \(\frac{11\; \times \;12 \;\times \;23}{6} = 506\)

Hence, **option 1** is the correct answer.