\sum_{k=1}^n4k^2

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∑_k=1ⁿ4k²

Solution

23 n(n+1)(2n+1)

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∑_k=1ⁿ4k²

Apply the constant multiplication rule: ∑c·a_n=c·∑a_n

=4· ∑_k=1ⁿk²

∑_k=1ⁿk²=16 n(n+1)(2n+1)

=4· 16 n(n+1)(2n+1)

Simplify 4· 16 n(n+1)(2n+1): 23 n(n+1)(2n+1)

=23 n(n+1)(2n+1)

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+▭▭	>	4	5	6	×	☐☐☐
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▭ \|▭	)	.	0	=	+	`y`

\sum_{k=1}^n4k^2

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