Track / Stage 5 / 361 of 400 #961 of 1964
Problem 961 AIME late Difficulty 5.9 Prove it
− 2 2 + 4 2 + 6 2 + ⋯ + ( 2 n ) 2 = 2 3 n ( n + 1 ) ( 2 n + 1 ) . \begin{array}{c}-2^{2}+4^{2}+6^{2}+\cdots+(2 n)^{2} \\ =\frac{2}{3} n(n+1)(2 n+1) .\end{array} − 2 2 + 4 2 + 6 2 + ⋯ + ( 2 n ) 2 = 3 2 n ( n + 1 ) ( 2 n + 1 ) .
The translation is as follows:
− 2 2 + 4 2 + 6 2 + ⋯ + ( 2 n ) 2 = 2 3 n ( n + 1 ) ( 2 n + 1 ) . \begin{array}{c}-2^{2}+4^{2}+6^{2}+\cdots+(2 n)^{2} \\ =\frac{2}{3} n(n+1)(2 n+1) .\end{array} − 2 2 + 4 2 + 6 2 + ⋯ + ( 2 n ) 2 = 3 2 n ( n + 1 ) ( 2 n + 1 ) .
This equation represents the sum of the squares of even numbers from − 2 2 -2^2 − 2 2 ( 2 n ) 2 (2n)^2 ( 2 n ) 2 2 3 n ( n + 1 ) ( 2 n + 1 ) \frac{2}{3} n(n+1)(2 n+1) 3 2 n ( n + 1 ) ( 2 n + 1 )
This one wants a proof. Work it on paper, then read the official solution and mark
yourself. Be honest about it: the record is only any use to you if it is.
I solved it I didn't Skip
Official solution \begin{array}{l}\text { Prove: } 2^{2}+4^{2}+6^{2}+\cdots+(2 n)^{2} \\ =2^{2} \cdot 1^{2}+2^{2} \cdot 2^{2}+2^{2} \cdot 3^{2}+\cdots \\ +2^{2} \cdot n^{2} \\ =2^{2}\left(1^{2}+2^{2}+3^{2}+\cdots+n^{2}\right) \\ =4 \cdot \frac{1}{6} n(n+1)(2 n+1) \\ =\frac{2}{3} n(n+1)(2 n+1) \text {. } \\\end{array}
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