Find the sum to n terms of the series : `5 + 11 + 19 + 29 + 41 dot dot dot`

Answer

N/a

Explanation

Difference of consecutive terms of the series (11-5), (19-11), (29-19),…
=6, 8, 10,… which are in A.P.
Therefore, `T_(n)` can be determined by the difference method.
`"Let "S_(n)= "sum of n terms."`
`S_(n)=5+11+19+29+41+...T_(n-1)+T_(n)`
`underline(S_(n)=" "5+11+19+29+.......+T_(n-1)+T_(n))`
On subtracting
`0=[5+6+8+10+12+.....+"nth terms"]-T_(n)`
`rArrT_(n)=5+[6+8+10+12+.....+"upto (n-1) terms"]`
`=5+(n-1)/(2)[2xx6+(n-1-1)xx2]`
`=5+(n-1)/(2)(2n+8)`
`=5+(n-1)(n+4)`
`=5+n^(2)+3n-4`
`=n^(2)+3n+1`
`S_(n)=Sigman^(2)+3Sigman+n`
`(1)/(6)n(n+1)(2n+1)+(3)/(2)n(n+1)+n`
`=(n(n+1)(2n+1)+9n(n+1)+6n)/(6)`
`=(1)/(6)n[(n+1)(2n+1)+9(n+1)+6]`
`=(1)/(6)n(2n^(2)+3n+1+9n+9+6)`
`=(1)/(6)n(2n^(2)+12n+16)`
`=(1)/(3)n(n^(2)+6n+8)`
`=(1)/(3)n(n+2)+(n+4).`