Evaluate the Sum 9Σ k = 1 k(9k + 7)?
Evaluate the Sum 9Σ k = 1 k(9k + 7)
the 9 is above the sigma and the k = 1 is below it.
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Evaluate the Sum 9Σ k = 1 k(9k + 7)
the 9 is above the sigma and the k = 1 is below it.
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sum(k = 1 to 9) 9k^2 + 7k = 9(sum k^2) + 7sum(k)
recall the sum (i = 1 to n) i^2 = [n(n + 1)(2n + 1)]/6 and sum(i = 1 to n) i = (n(n + 1))/2
9((9)(10)(19)/6) + 7(9)(10)/2
2565 + 315 = 2880