Question
Download Solution PDFWhat is the value of 22 + 42 + 62 + …… + 202?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept 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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