Question

Find the sum to `n` terms of the series `3+15+35+63+`

Answer 1

Correct Answer - `(1)/(3)n(4n^(2)+6n-1)`
`S_(n)=3+15+35+63+ ...+T_(n-1)+T_(n)" " `...(i)
`S_(n)=3+15+35+ ...+T_(n-2)+T_(n-1)+T_(n). " " `...(ii)
On subtracting (ii) from (i), we get
`0=3+[12+20+28+..."to"(n-1)"terms"]-T_(n)`
`rArr T_(n)=3+(1)/(2)(n-1){2 xx12+(n-2)xx8}=(4n^(2)-1)`
`rArr S_(n) =4(sum_(k=1)^(n)k^(2))-n={4xx(1)/(6)n(n+1)(2n+1)}-n=(1)/(3)n(4n^(2)+6n-1)`.