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Question
Find the sum of the following series
1 + 4 + 9 + 16 + ... + 225
Solution
1 + 4 + 9 + 16 + ... + 225
= 12 + 22 + 32 + 42 + ... + 152
`sum_1^"n" "n"^2 = ("n"("n" + 1)(2"n" + 1))/6`
`sum_1^15 "n"^2 = (15(15 + 1)(2 xx 15 + 1))/6`
= `(15 xx 16 xx 31)/6`
= 1240
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