Find the sum of n terms of the series 1+4+10+20+35+…

Answer

N/a

Explanation

Let nth term and sum of n terms of the series be `T_(n)" and "S_(n)` respectively.
`S_(n)=1+4+10+20+35+...+T_(n-1)+T_(n)`
`underline(S_(n)=" "1+4+10+20+...+T_(n-1)+T_(n))`
On subtracting
`0=1+3+6+10+15+...+t_(n)-T_(n)`
`rArr" "T_(n)=1+3+6+10+15+...+t_(n)`
`underline(T_(n)=" "1+3+6+10+...+t_(n-1)+t_(n))`
On subtracting
`0=[1+2+3+4+5+..."nth term"]-t_(n)`
`rArr" "t_(n)=Sigman=(n(+1))/(2)`
`=(1)/(2)(n^(2)+n)`
`rArr" "T_(n)=(1)/(2)(Sigman^(2)+Sigman)`
`=(1)/(2)[(1)/(6)n(n+1)(2n+1)+(1)/(2)n(n+1)]`
`=(1)/(2)n(n+1)[(2n+1+3)/(6)]`
`=(1)/(6)n(n+1)(n+2)`
`rArr" "T_(n)=(1)/(6)[n^(2)+3n^(2)+2n]`
`rArr" "S_(n)=(1)/(6)[Sigman^(2)+3Sigman^(2)+2Sigman]`
`=(1)/(6)[(1)/(4)n^(2)(n+1)^(2)+(3)/(6)n(n+1)(2n+1)+(2)/(2)n(n+1)]`
`=(1)/(6)n(n+1)[(1)/(4)n(n+1)+(1)/(2)(2n+1)+1]`
`=(1)/(6)n(n+1)[(n(n+1)+2(2n+1)+4)/(4)]`
`=(1)/(24)n(n+1)(n^(2)+n+4n+2+4)`
`=(1)/(24)n(n+1)(n^(2)+5n+6)`
`=(1)/(24)n(n+1)(n+2)(n+3)`