NIMCET 2011 — Mathematics PYQ

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Question

NIMCET 2011 — Mathematics PYQ

NIMCET | Mathematics | 2011

The sum of $11^2 + 12^2 + \dots + 30^2$ is:

Choose the correct answer:

Explanation

1. Recall the Formula:

The sum of squares of the first $n$ natural numbers is given by:

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

2. Set up the Calculation:

The required sum can be calculated as the sum of squares up to $30$ minus the sum of squares up to $10$ :

\text{Sum} = (1^2 + 2^2 + \dots + 30^2) - (1^2 + 2^2 + \dots + 10^2)

\text{Sum} = \sum_{k=1}^{30} k^2 - \sum_{k=1}^{10} k^2

3. Calculate $\sum_{k=1}^{30} k^2$ (where $n=30$ ):

\frac{30(30+1)(2 \times 30 + 1)}{6} = \frac{30 \times 31 \times 61}{6}

= 5 \times 31 \times 61 = 9455

4. Calculate $\sum_{k=1}^{10} k^2$ (where $n=10$ ):

\frac{10(10+1)(2 \times 10 + 1)}{6} = \frac{10 \times 11 \times 21}{6}

= \frac{2310}{6} = 385

5. Find the final difference:

\text{Sum} = 9455 - 385

\text{Sum} = 9070

Conclusion:

The sum of the series is $9070$ .

The correct option is (b).