Alg8.11

Find the sum to $n\,$ terms of the series $1\cdot3+3\cdot5+5\cdot7+........\,$

If we write the first factors of each term in the same order $1,3,5......\,$ which form an AP.

The nth term of the series is $1+(n-1)2=2n-1\,$

Similarly nth term of the second factors is $2n+1\,$

The nth term of the given series is $(2n-1)(2n+1)=4n^2-1\,$

Sum is $S_n=\sum (4n^2-1)=4\sum n^2-\sum 1\,$

$S_n=4\cdot \frac{n(n+1)(2n+1)}{6}-n\,$

$S_n=\frac{2n(n+1)(2n+1)-3n}{3}=\frac{n(4n^2+6n-1)}{3}\,$