# Induction and formulating equations

• Sep 18th 2010, 11:32 PM
jspstorm
Induction and formulating equations
While browsing, I came across equations for finding the sums of certain series's, the first is (n²+n)/2 for "1+2+3...+n", the second is (1/6)n(n+1)(2n+1) for "1²+2²+3²...n²".

I found out how to come up with the first equation through reasoning: n(n+1)/2 is basically the median times n, and since the difference in this series is 1, and the series begins with 1, the median is the same as the mean, so the mean times the number of numbers in the series finds the total.

Now what I can't figure out is how to formulate the second equation, any ideas?
• Sep 18th 2010, 11:49 PM
Prove It
• Sep 20th 2010, 08:17 AM
jspstorm
Why the 3x multiplier in 3(1²+2²+...+n²)?
• Sep 20th 2010, 01:28 PM
emakarov
The shaded area on the right is equal to the total area of the shaded squares on the left (small squares with the same type of shading from various big squares are collected together). Also, in the picture on the right, the blank area to the left of the shaded figure is equal to 1*1 + 2*2 + 3*3 + 4*4 + 5*5: look at it bottom up. The same goes for the blank area right of the shaded figure. Thus, the total area of the rectangle on the right is three times the total area of the squares on the left.
• Sep 20th 2010, 05:28 PM
Prove It
And you already know that $1 + 2 + 3 + \dots + n = \frac{1}{2}n(n+ 1)$.

So that means

$3(1^2 + 2^2 + 3^2 + \dots + n^2) = (1 + 2 + 3 + \dots + n)(2n + 1)$

$= \frac{1}{2}n(n + 1)(2n + 1)$.

Thus

$1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{1}{6}n(n + 1)(2n + 1)$.
• Sep 21st 2010, 04:10 AM
Quote:

Originally Posted by jspstorm
While browsing, I came across equations for finding the sums of certain series's, the first is (n²+n)/2 for "1+2+3...+n", the second is (1/6)n(n+1)(2n+1) for "1²+2²+3²...n²".

I found out how to come up with the first equation through reasoning: n(n+1)/2 is basically the median times n, and since the difference in this series is 1, and the series begins with 1, the median is the same as the mean, so the mean times the number of numbers in the series finds the total.

Now what I can't figure out is how to formulate the second equation, any ideas?

You could also create some numerical patterns as follows...

$1^2+2^2+3^2+4^2+5^2+...+n^2=1+4+9+16+25+...+n^2$

$=1+(1+3)+(1+3+3+2)+(1+3+3+3+2+2+2)+(1+3+3+3+3+2+2+ 2+2+2+2)+....$

$=1[n]+3[0+1+2+3+4+5+...+(n-1)]+2[0+0+1+3+6+10+15+21+...]$

Two of these are very straightforward to evaluate.

1 appears n times.
3 appears in (n-1) brackets and the sum is a simple arithmetic series of (n-1) terms.

2 appears in (n-2) brackets and the sum is the sum of triangular numbers
whose sequence can be seen on the 3rd diagonal from the top of Pascal's triangle.
The sum of triangular numbers can be seen on the diagonal below that (the 4th from the top).

$1+3=4,\;\;4+6=10,\;\;10+10=20,\;\;20+15=35......$

These sums are $\binom{n+2}{3}$

As there are (n-2) terms in our sum, the sum within brackets of the 2's is $\binom{n}{3}$

Hence

$\displaystyle\ 1^2+2^2+3^2+...+n^2=n+3\frac{(n-1)(n)}{2}+2\frac{n(n-1)(n-2)}{3!}$

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

$=\displaystyle\frac{6n+9n(n-1)+2n\left(n^2-3n+2\right)}{6}$

$=\displaystyle\frac{n\left(2n^2-6n+4+9n-9+6\right)}{6}$

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