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

Aug 12, 2016

$650$.

Explanation:

We have the formula

${\sum}_{i = 1}^{i = n} {i}^{2} = {1}^{2} + {2}^{2} + {3}^{2} + \ldots + {n}^{2} = \frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right)$.

$\therefore {\sum}_{i = 1}^{i = 12} {i}^{2} = {1}^{2} + {2}^{2} + {3}^{2} + \ldots + {12}^{2} = \frac{12}{6} \left(12 + 1\right) \left(25\right)$

$= 2 \cdot 13 \cdot 25 = 650$.

Aug 12, 2016

$650$

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:

$1 , 4 , 9 , 16 , 25 , \ldots$

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

$\textcolor{b l u e}{0} , 1 , 5 , 14 , 30 , 55 , \ldots$

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

$\textcolor{b l u e}{1} , 4 , 9 , 16 , 25 , \ldots$

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

$\textcolor{b l u e}{3} , 5 , 7 , 9 , \ldots$

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

$\textcolor{b l u e}{2} , 2 , 2 , \ldots$

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)

$= n + \frac{3}{2} {n}^{2} - \frac{3}{2} n + \frac{1}{3} {n}^{3} - {n}^{2} + \frac{2}{3} n$

$= \frac{1}{6} \left(2 {n}^{3} + 3 {n}^{2} + n\right)$

$= \frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right)$

So:

${S}_{12} = \frac{1}{6} \cdot 12 \cdot 13 \cdot 25 = 650$