Find the sum of the following series to `n` term: `3xx1^2+5xx2^2+7xx3^2+`

Answer

N/a

Explanation

Given series `3xx1^(2)+5xx2^(2)xx7xx3^(2)+……` nth term of given series.
=(nth term of 3+5+7+…)`xxn^(2)`
`=[3+(n-1)2]xxn^(2)=(2n+1)xxn^(2)`
`=2n^(3)+n^(2)`
`:.` Sum of n term of given series.
`S_(n)=sum(2n^(3)+n^(2))=2 sum^(3)+sumn^(2)`
`=2((n(n+1))/(2))^(2)+(n(n+1)(2n+1))/(6)`
`=(n(n+1))/(2)[2cdot(n(n+1))/(2)+(2n+1)/(3)]`
`=(n(n+1))/(2)[(3n(n+1)+2n+1)/(3)]`
`=(n(n+1))/(2)[(3n^(2)+5n+1)/(3)]`
`=(n(n+1)(3n^(2)+5n+1))/(6)`
Therefore, sum of n terms of the series
`=(n(n+1)(3n^(2)+5n+1))/(6)`