### 1 Answer | Add Yours

Let S = 2+7+14+23+............+Tn------(i) (Tn is the nth term)

again

S = 2+7+14+23+............+Tn-1+Tn -----(ii)

On subtracting Eq. (ii) from Eq. (i), we get

`=>` Tn = 2+5+7+9+....+(Tn-Tn-1)

= `2+(n-1)/2 (10+(n-2)*2)`

= 2+ (n-1)(n+3)

= `n^2+2n-1`

Hence S = `Sigma(n^2+2n-1)`

= `Sigman^2+2Sigman-Sigma1`

=`(n(n+1)(2n+1))/6 + (2n(n+1))/2-n`

=`(n(2n^2+3n+1+6n+6-6))/6`

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

Putting n=99,

S =` (99(2*99^2+9*99+1))/6`

**= 338151 => answer**

**Sources:**

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes