The sum to n terms of the series $3^2 + 8^2 + 13^2$ +.......
- $\dfrac{n(50n^2 + 15n-11)}{6}$
- $\dfrac{(50n^2+15n-11)}{6}$
- $\dfrac{n(50n^2+15n+11)}{6}$
- $\dfrac{n(50n^2-15n-11)}{6}$
The series is sum of (5k-3)^2 for k=1 to n. (5k-3)^2 = 25k^2 - 30k + 9. Sum = 25 * [n(n+1)(2n+1)/6] - 30 * [n(n+1)/2] + 9n. Simplifying this leads to n(50n^2 + 15n - 11) / 6.
Identify the base numbers of the series (3, 8, 13...) as an arithmetic progression with the first term a = 3 and common difference d = 5, making the nth base term 5n - 2. The sum of the squares of these terms is the sum of (5n - 2)^2 from 1 to n, which expands to the sum of (25n^2 - 20n + 4). Using the standard formulas for the sum of n natural numbers, the sum of their squares, and the sum of constants, evaluate 25 * n(n+1)(2n+1)/6 - 20 * n(n+1)/2 + 4n. Expanding and simplifying the expression results in n(50n^2 + 15n - 11) / 6.