Serious series

Algebra Level 3

1 ⋅ 2 2 + 2 ⋅ 3 2 + ⋯ + n ( n + 1 ) 2 1 2 ⋅ 2 + 2 2 ⋅ 3 + ⋯ + n 2 ( n + 1 ) = ? \dfrac{1\cdot 2^2 + 2\cdot 3^2 + \cdots + n(n+1)^2 }{1^2 \cdot 2 + 2^2 \cdot 3 + \cdots + n^2(n+1) } = \, ?

3 n + 1 3 n + 5 \frac{3n+1}{3n+5} 3 n − 5 3 n − 1 \frac{3n-5}{3n-1} 3 n + 5 3 n + 1 \frac{3n+5}{3n+1} 3 n + 2 3n+2

Guneet Singh Oberai by Guneet Singh Oberai , 23, India

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1 solution

Substitute n=0 into the equation. The answer is hence found.

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