Question

Find the sum of n terms of the series 1^2 + 3^2 + 5^2 + 7^2+ . . . . . . . .​

Answer 1

TO FIND

Sum of n terms of the series 1² + 3² + 5² + 7² + . .

SOLUTION

Series ➠‎ 1² + 3² + 5² . . . . (2n - 1)²

Let Tn be nth term of the series.

$o, Tn = (2n−1)² = 4n² − 4n + 1

∑Tn =∑4n² − 4n + 1

= 4∑n² −4∑n + ∑1

• ∑n² = n(n + 1)(2n + 1)/4
• ∑n = n(n+1)/4
• ∑ = Sign of summation

= 4 × [n(n+1)(2n+1)/4] - 4 × [n(n+1)/2] + n

= n(n+1) [2/3 (2n + 1) - 2] + n

= n(n+1) [(4n + 2 - 6)/3] + n

= n/3 (n + 1) (4n - 4) + n

= 4n/3 (n + 1)(n - 1) + n

= 4/3 n(n² - 1) + n

= n/3 (4n² - 4 + 3)

= n(2n - 1)(2n + 1) / 3

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Answer 2

Answer:

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