The sum of the series ${1}^{2}+{3}^{2}+{5}^{2}+...$ to $n$ terms is
- $\cfrac{n(n+1)(n+!)}{2}$
- $\cfrac{n(2n-1)(2n+1)}{3}$
- $\cfrac{{(n-1)}^{2}(2n+1)}{6}$
- $\cfrac{{(2n+1)}^{3}}{3}$
Reveal answer
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B
Correct answer
Explanation
The kth odd number is 2k - 1, so the required sum is the sum of (2k - 1)^2 from k = 1 to n. Its closed form is n(2n - 1)(2n + 1)/3, which is option B.
AI explanation
The nth term of the series is the square of the nth odd number, which can be written as (2n - 1)^2 = 4n^2 - 4n + 1. We find the sum to n terms by applying the standard formulas for the sum of n natural numbers and the sum of n squared natural numbers, resulting in 4 * n(n+1)(2n+1)/6 - 4 * n(n+1)/2 + n. Multiplying the entire expression by 3n to clear the denominator gives n(2n+1)(2n+2) - 6n(n+1) + 3n, which factors perfectly into n(2n+1)(2n-1)/3. The result is n(2n-1)(2n+1)/3.