Homework Help

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

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christiano-cr7 | (Level 1) Salutatorian

Posted July 29, 2013 at 4:55 AM via web

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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?

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted July 29, 2013 at 6:14 AM (Answer #1)

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`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.

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