relation between arithmetic series and `square` arithmetic series

Question

For example:

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

$$1^2+2^2+\text{...}+n^2=\frac{n(n+1)(2n+1)}{6}~~~(2)$$

In this equality, I sometimes recall by heart

$\frac{n(2n+1)(2n+3)}{6}$ or others.

Why I cannot memorize some formulas exactly over these years?

Question1

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How to derive $\frac{n(n+1)(2n+1)}{6}$ from $1^2+2^2+\text{...}+n^2$?

Question2

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Now I happen to notice that $n(n+1)$ are the same to the summation in

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

What's the relation between (1) and (2).

I think there must be large materials about more general cases

$$\sum _{i=0}^n i^k\text{ },k=1,2,\text{...},n~~~(3)$$

What's the formal terminology about that?

Answer 1

First, we look at your observation that $$1+2+3+\cdots+n=\frac{n(n+1)}{2}.\tag{1}$$ If you are somewhat familiar with combinatorics, there is a nice way of seeing it via the binomial coefficient $\binom{n+1}{2}$, sometimes called $C(n+1,2)$ or $C_2^{n+1}$ (there are other names).

The right-hand side of (1) is the number of ways to choose $2$ numbers from the $n+1$ numbers $1$ to $n+1$. Let's count the number of ways to choose $2$ numbers in another way.

Maybe the second largest number chosen is $n$. There is then $1$ way of choosing the largest number.

Maybe the second largest number chosen is $n-1$. There are then $2$ ways to choose the largest number.

Maybe the second largest number chosen is $n-2$. There are then $3$ ways to choose the largest number.

Continue. Finally, if the second largest number chosen is $1$, there are $n$ ways to choose the largest number.

It follows that $\binom{n+1}{2}=1+2+3+\cdots +n$.

The nice thing about this viewpoint is that it generalizes. The same reasoning shows, for example, that $$\binom{2}{2}+\binom{3}{2}+\binom{4}{2}+\cdots+\binom{n}{2} =\binom{n+2}{3}.\tag{2}$$

The binomial coefficient on the right of (2) is equal to $\frac{(n+2)(n+1)(n)}{3!}.$

The sum on the left of (2) is $$\frac{(1)(2)}{2}+\frac{(2)(3)}{2}+\frac{(3)(4)}{2}+\cdots +\frac{n(n+1)}{2}.\tag{3}.$$ The sum (3) is a close relative of the sum of the first $n$ squares. For it is equal to $$\frac{1}{2}\left(1^2+1+2^2+2+3^2+3+\cdots+n^2+n\right).$$ Since we already know the sum $1+2+3+\cdots +n$, we can now find a formula for the sum of the first $n$ squares. Putting things together, it is $$2\frac{(n+2)(n+1)(n)}{3!} -\frac{n(n+1)}{2}.$$ Algebraic simplification (bring to common denominator $6$) then yields the hard to remember formula for the sum of the first $n$ squares.

Answer 2

Wikipedia has a good derivation of the formula for squares:http://en.wikipedia.org/wiki/Square_pyramidal_number

You're right that the formulas are connected. The pattern that connects them is called the Bernoulli formula, which uses Bernoulli numbers. They were invented to study sums of powers, and they are connected to incredibly beautiful areas of number theory and other areas of mathematics.

Answer 3

You are looking for Faulhaber's formulas

Answer 4

As $(r+1)^{k+1}-r^{k+1}=\binom{k+1}1r^k+ \binom{k+1}2r^{k-1}+\cdots+\binom{k+1}{k-1}r^2+\binom{k+1}kr+1$

Putting $r=1,2,3,\cdots,n-1,n$ and adding them we get

$$(n+1)^{k+1}-1=\binom{k+1}1S_k+\binom{k+1}2S_{k-1}+\cdots+\binom{k+1}{k-1}S_2+\binom{k+1}kS_1+S_0 $$ where $S_k=\sum_{1\le r\le n}r^k$

Now, $S_0=\sum_{1\le r\le n}1=n$

Putting $k=1,(n+1)^2-1=\binom21S_1+S_0$ which will give us $S_1$ as we know $S_0$

Putting $k=2,(n+1)^3-1=\binom31S_2+\binom32S_1+S_0$ which will give us $S_2$ as we know $S_0,S_1$

and so on