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

Question

Lily Anderson · Accepted Answer

#650#.

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#.

Elias Ackerman · Answer

#650#

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#

Theodore Amidon · Answer

undefined To find the sum of the series $i^2$ from $i=1$ to $12$, you can use the formula for the sum of the first $n$ perfect squares, which is $\frac{n(n+1)(2n+1)}{6}$. Substituting $n=12$ into this formula, you get:

$$\frac{12(12+1)(2*12+1)}{6}$$

Simplify the expression:

$$\frac{12(13)(25)}{6}$$

$$\frac{3900}{6}$$

$$650$$

So, the sum of the series $i^2$ from $i=1$ to $12$ is $650$.