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Algebra Difficulty 5.4 AIME, harder Prove it

 Eight, 1⋅3⋅5+3⋅5⋅7+⋯+(2n−1)(2n+1)(2n+3)=n(2n3+8n2+7n−2).\begin{array}{l}\text { Eight, } 1 \cdot 3 \cdot 5+3 \cdot 5 \cdot 7+\cdots+(2 n-1)(2 n \\ \quad+1)(2 n+3)=n\left(2 n^{3}+8 n^{2}+7 n-2\right) .\end{array}

Solution

Using the general formula an=(2n−1)(2n+1)(2na_{n}=(2 n-1)(2 n+1)(2 n +3)=8n3+12n2−2n−3+3)=8 n^{3}+12 n^{2}-2 n-3, we can prove by analogy with formula six.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.