श्रेणी `1+4+10+20+35+….` का n पदों तक योगफल ज्ञात कीजिए।

Explanation

माना श्रेणी का n वां पद `T _(n )` और n पदों का योगफल `S _(n )` है।
`{:(" "S_(n)=1+4+10+20+35+...+T_(n-1)+T_(n)),(" "ul(S_(n)= " "1+4+10+20+...+T_(n-1)+T_(n))),("घटाने पर"),(" "0=1+3+6+10+15+...+t_(n)-T_(n)):}`
`{:(" "rArr T_(n)=1+3+6+10+15+...+t_(n)),(" "ul(T_(n)=" "1+3+6+10+...+t_(n-1)+t_(n))),("घटाने पर"),(" "0=[1+2+3+4+5+...n "वां पद"]-t_(n)):}`
`rArr t_(n)=Sigman=(n(n+1))/(2)=(1)/(2)(n^(2)+n)`
`rArr T_(n)=(1)/(2)(Sigma n^(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^(3)+3n^(2)+2n]`
`rArr S_(n)=(1)/(6)[Sigma n^(3)+3Sigma n^(2)+2 Sigma n]`
`=(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)+2n(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)`