The number pattern 1, 5, 11, 19, … is such that the sequence of 'second differences' is a constant.

What is the sum of 100 numbers in the pattern?

### 1 Answer | Add Yours

`T_1 = 1`

`T_2 = 5 = 2^2+1`

`T_3 = 11 = 3^2+2`

`T_4 = 19 = 4^2+3`

So using the pattern we can say that the nth term of the series would be;

`T_n = n^2+(n-1)`

We need to find sum of 100 terms. So n = 100

`S_n=sum_(r = 1)^nT_r`

`S_n = sum_(r=1)^100(r^2+r-1)`

`S_n = sum_(r=1)^100(r^2)+sum_(r=1)^100(r)-sum_(r=1)^n(1)`

`sum_(r=1)^n(r^2) = (n(n+1)(2n+1))/6`

`sum_(r=1)^n(n) = (n(n+1))/2`

`sum_(r=1)^n(1) = n`

`S_n = sum_(r=1)^100(r^2)+sum_(r=1)^100(r)-sum_(r=1)^n(1)`

`S_100 = (100(100+1)(2xx100+1))/6+(100(100+1))/2-100`

`S_100 = 338350+5050-100`

`S_100 = 343300`

*So the sum of 100 terms in the series is 343300.*

**Sources:**

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes