Find the nth term and sum to n tems of the following series: 1+5+12+22+…………..

Explanation

The sequence of differences between successuve terms is `4,7,10,13,"….",`.Clearly, it is an AP is in common difference 3. S0, let the nth rerm of the given series be `T_(n)` and sum of n temrs be `S_(n)`.
Then, `S_(n)=1+5+12+22+35+"..."+T_(n-1)+T_(n) "...(i)"`
`S_(n)=1+5+12+22+"..."+T_(n-1)+T_(n) "...(ii)"`
Subtracting Eq. (ii) from Eq. (i), we get
`0=1+4+7+10+13+"..."+(T_(n)-T_(n-1))-T_(n)`
`implies T_(n)=1+4+7+10+13+"..." " terms "`
`(n)/(2){2*1+(n-1)3}=(1)/(2)(3n^(2)-n)`
Hence, `T_(n)=(3)/(2)n^(2)-(1)/(2)n`
`therefore` Sum of n terms `S_(n)=sum T_(n)=(3)/(2)sumn^(2)-(1)/(2)sumn`
`=(3)/(2)((n(n+1)(2n+1))/(6))-(1)/(2)(n(n+1))/(2)`
`=(n(n+1))/(4)(2n+1-1)`
`=(1)/(2)n^(2)(n+1)=(1)/(2)(n^(3)+n^(2))`