An algebra problem by Batman

Algebra Level 3

The first and last terms of an arithmetic progression are a a and b b respectively. There are altogether ( 2 n + 1 ) (2n+1) terms. A new series is formed by multiplying each of the first 2 n 2n terms by the next term. Find the sum of this new series.

( 4 n − 1 ) ( a 2 + b 2 ) + ( 4 n + 2 ) a b 6 n \frac{(4n-1)(a^2+b^2)+(4n+2)ab}{6n} ( 4 n − 1 ) ( a 2 + b 2 ) + ( 4 n 2 + 2 ) a b 6 n \frac{(4n-1)(a^2+b^2)+(4n^2+2)ab}{6n} ( 4 n 2 − 1 ) ( a 2 + b 2 ) + ( 4 n 2 + 2 ) a b 6 n \frac{(4n^2-1)(a^2+b^2)+(4n^2+2)ab}{6n} ( 4 n 2 − 2 ) ( a 2 + b 2 ) + ( 4 n 2 + 1 ) a b 6 n \frac{(4n^2-2)(a^2+b^2)+(4n^2+1)ab}{6n}

Rudraksh Shukla by Rudraksh Shukla , 23, India

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