Find the sum of nth term of the series `=1+4+10+20+35+".."`.

Explanation

The sequence of first consctive differences is `3,6,10,15,"…"` and second consecutive differences is `3,4,5,"….",`. Clearly, it is an AP with comon difference 1. So, let the nth term and sum of the seies upto n terms of the series be `T_(n)` and `S_(n)`, respectively.
Then, `S_(n)=1+4+10+20+35+".."+T_(n-1)+T_(n)"....(i)"`
`S_(n)=1+4+10+20+".."+T_(n-1)+T_(n)"....(ii)"`
Subtracting Eq. (ii) from Eq.(i), we get
`0=1+3+6+10+15+".."+(T_(n)-T_(n))-T_(n)`
`implies T_(n)=1+3+6+10+15+".." " upto n terms "`
or ` T_(n)=1+3+6+10+15+".."+t_(n-1)+t_(n)"....(iii)"`
` T_(n)=1+3+6+10+".."+t_(n-1)+t_(n)"....(iv)"`
Now, subtracting Eq. (iv) from Eq. (iii), we get ` 0=1+2+3+4+5+".."+(t_(n)-t_(n-1))-t_(n)`
or `t_(n)=1+2+3+4+5+".." " upto n terms "`
`=sum n=(n(n+1))/(2)`
` therefore T_(n)=sumt_(n)=(1)/(2)(sumn^(2)+sumn)`
`(1)/(2)((n(n+1)(2n+1))/(2)+(n(n+1))/(2))`
`(1)/(2)*(n(n+1))/(6)(2n+1+3)=(1)/(6)n(n+1)(n+2)`