Question

How do you find the sum of the series `i^2` from i=1 to 12?

Answer 1

`650`.

Explanation:

We have the formula

`sum_(i=1)^(i=n) i^2=1^2+2^2+3^2+...+n^2=1/6n(n+1)(2n+1)`.

`:. sum_(i=1)^(i=12)i^2=1^2+2^2+3^2+...+12^2=12/6(12+1)(25)`

`=2*13*25=650`.

Answer 2

`650`

Explanation:

Here's a non-standard way to do it without having to remember individual formulae for different kinds of sequences...

The sequence of squares looks like this:

`1, 4, 9, 16, 25,...`

So the sequence of sums of squares from `0` starts like this:

`color(blue)(0), 1, 5, 14, 30, 55,...`

The sequence of differences of that sequence is the sequence of squares:

`color(blue)(1), 4, 9, 16, 25,...`

The sequence of differences of the sequence of squares is a linear (arithmetic) sequence:

`color(blue)(3), 5, 7, 9,...`

The sequence of differences of that sequence is a constant sequence:

`color(blue)(2), 2, 2,...`

We can then use the first term of each of these sequences as coefficients to give a formula for the sum of `n` terms of the sequence of squares:

`S_n = color(blue)(0)/(0!) + color(blue)(1)/(1!) n + color(blue)(3)/(2!) n(n-1) + color(blue)(2)/(3!) n(n-1)(n-2)`

`=n + 3/2n^2-3/2n+1/3n^3-n^2+2/3n`

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

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

So:

`S_12 = 1/6*12*13*25 = 650`