Find the sum of first `n` terms of the following series:`5+11+19+29+41+......`

Explanation

It is not an AP or a GP

Let
`S_n=5+11+19+29+41+......+a_(n-1)+a_n " " " " .....(1)`

`S_n=0+5+11+19+29+41+......+a_(n-2)+a_(n-1)+a_n " " " " .....(2)`

Subtracting `(2)` from `(1)`

`S_n-S_n=5-0+[(11-5)+(19-11)+(29-19)+.....(a_(n-1)-a_(n-2))+(a_n-a_(n-1))]-a_n`

`0=5+[6+8+10+12+...a_(n-1)]-a_n`

`a_n=5+[6+8+10+12+...+(n-1)`terms`] " " " " ....(3)`

`6+8+10+12+...+(n-1)`term is an AP

With first term, `a=6` & Common Difference,`d=8-6=2`

Sum of `n` terms of an Ap `=n/2(2a+(n-1)d)`

Putting `n=n-1` & `a=6` & `d=2`

`[6+8+10+12+...+(n-1)`terms`]=(n-1)/2[2(6)+((n-1)-1)2]`

` " " " "=(n-1)/2[12+(n-1-1)2]`

` " " " "=(n-1)/2[12+(n-2)2]`

` " " " "=(n-1)/2[12+2n-4]`

` " " " "=(n-1)/2[8+2n]`

` " " " "=(n-1)/2xx2[4+n]`

` " " " "=(n-1)(n+4)`

Thus, `[6+8+10+12+...+(n-1)`terms`]=(n-1)(n+4)`

Now ,
`a_n=5+[6+8+10+12+...+(n-1)`terms`]`

Putting values

`a_n=5+(n-1)(n+4)`

`a_n=5+n(n+4)-1(n+4)`

`a_n=5+n^2+4n-n-4`

`a_n=n^2+3n+1`

Now,

`S_n=sum_(n=1)^(n)a_n`

`S_n=sum_(n=1)^(n)n^2+3n+1`

`S_n=sum_(n=1)^(n)n^2+sum_(n=1)^(n)3n+sum_(n=1)^(n)1`

`S_n=sum_(n=1)^(n)n^2+3sum_(n=1)^(n)n+sum_(n=1)^(n)1`

` " " " " =(n(n+1)(2n+1))/6+3((n(n+1))/2)+n`

` " " " " =(n(n+1)(2n+1))/6+3/2n(n+1)+n/1`

` " " " " =(n(n+1)(2n+1)+9n(n+1)+6n)/6`

` " " " " =n(((n+1)(2n+1)+9(n+1)+6)/6)`

` " " " " =n((2n^2+12n+16)/6)`

` " " " " =n((2(n^2+6n+8))/6)`

` " " " " =n/3(n^2+6n+8)`

` " " " " =n/3[n(n+4)+2(n+4)]`

` " " " " =n/3[(n+2)(n+4)]`

` " " " " =[n(n+2)(n+4)]/3`


Thus, the required sum is `[n(n+2)(n+4)]/3`