What is the value of 22 + 42 + 62 + …… + 202?

This question was previously asked in
OSSTET 2018 (Science PCM) Official Paper (Held on 5 Oct 2018)
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  1. 770
  2. 1155
  3. 1540
  4. 385 × 385

Answer (Detailed Solution Below)

Option 3 : 1540
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Detailed Solution

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Concept Used:

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

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

Calculation:

22 + 42 + 62 + …… + 202 = 22 + (2 × 2)2 + (2 × 3)2 . . . + (2 × 10)2

⇒ 22 × (12 + 22 + 32 + …… + 102)

⇒ 4 ×\( \frac{10 (10+1)(2×10+1)}{6}\) = 4 × \( \frac{10 ×11 × 21 }{6}\)

⇒ 2 × 10 × 11 × 7 = 1540 

Hence, the required value is 1540.

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