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

Explanation

The difference between the successive terms are 15-3=12,35-15=20,63-35=28,….. Clearly, these differences are in A.P. Let `T_(n)` be the nth term and `S_(n)` denote the sum to n terms of the given series. Then,
`{:(S_(n)=3+15+35+63+...+T_(n-1)+T_(n)," "(1)),(S_(n)=" "3+15+35+63+...+T_(n-1)+T_(n)," "(2)),(bar(0=3+[12+20+28+....+(n-1)"terms"]-T_(n)),):}`
[Subtracting (2) from (1)]
`rArrT_(n)=3+((n-1))/2[2xx12+(n-1-1)xx8]`
`=3+(n-1)(12+4n-8)`
`=3+(n-1)(4n+4)`
`=4n^(2)-1`
`rArrS_(n)=sum_(k=1)^(n)T_(k)=sum_(k=1)^(n)(4k^(2)-1)=4sum_(k=1)^(n)k^(2)-sum_(k=1)^(n)1`
`=4{(n(n+1)(2n+1))/6}-n`
`n/3(4n^(2)+6n-1)`
Alternative method:
When the differences of consecutive terms are in A.P., nth term is given by,
`T_(n)=an^(2)+bn+c`
Here, `T_(1)=a+b+c=3` (1)
`T_(2)=4a+2b+c=15` (2)
`T_(3)=9a+3b+c=35` (3) From (2)-(1), we get
3a+b=12 (4)
From (3)-(2), we get
5a+b=20 (5)
From (5)-(4), we get
2a=8 or a=4
From (4), b=0
From (1), c=-1
So, `T_(n)=4n^(2)-1`
`thereforeS_(n)=sum_(k=1)^(n)T_(k)=n/3(4n^(2)+6n-1)`