We'll be looking into a methdo of getting a closed-form formulas for

\displaystyle\sum_{i=1}^{n}{i}^{k}.

The basis of the derivation is in the following representation

{1}^{k}+2^{k}+\cdots+{n}^{k}= {1}^{k-1}+2^{k-1}+3^{k-1}+\cdots+{n}^{k-1}
           +2^{k-1}+3^{k-1}+\cdots+{n}^{k-1}
                      +3^{k-1}+\cdots+{n}^{k-1}
                        \cdots

From which it follows that

\displaystyle\sum_{i=1}^{n}{i}^{k} = \displaystyle\sum_{j=1}^{n}\displaystyle\sum_{i=j}^{n}{i}^{k-1}
 =\displaystyle\sum_{j=1}^{n}(\displaystyle\sum_{i=1}^{n}{i}^{k-1} + ({j}^{k-1} - \displaystyle\sum_{i=1}^{j}{i}^{k-1}))
 =n\displaystyle\sum_{i=1}^{n}{i}^{k-1} + \displaystyle\sum_{j=1}^{n}{j}^{k-1} - \displaystyle\sum_{j=1}^{n}\displaystyle\sum_{i=1}^{j}{i}^{k-1}

Therefore, for k\ge{1},

\displaystyle\sum_{i=1}^{n}{i}^{k} = (n+1)\displaystyle\sum_{i=1}^{n}{i}^{k-1}-\displaystyle\sum_{j=1}^{n}\displaystyle\sum_{i=1}^{j}{i}^{k-1}

For example, for k=1,

\displaystyle\sum_{i=1}^{n}{i} = (n+1)\displaystyle\sum_{i=1}^{n}{1} - \displaystyle\sum_{j=1}^{n}{j}

which implies that

\displaystyle\sum_{i=1}^{n}{i} = \frac{(n+1)n}{2}.

For k=2 we get

\displaystyle\sum_{i=1}^{n}{i}^{2}

= (n+1)\displaystyle\sum_{i=1}^{n}{i}-\displaystyle\sum_{j=1}^{n}\displaystyle\sum_{i=1}^{j}{i}

= \frac{(n+1)^{2}{n}}{2} -\frac{1}{2}\displaystyle\sum_{j=1}^{n}({j}^{2}+{j})

from which it follows that

\frac{3}{2}\displaystyle\sum_{i=1}^{n}{i}^{2}=\frac{{n}^{2}(n+1)}{2} - \frac{n(n+1)}{4}

so that

\displaystyle\sum_{i=1}^{n}{i}^{2}=\frac{n(n+1)(2n+1)}{6}

Here is a list of a few additional sums:

\displaystyle\sum_{i=1}^{n}{i}^{3} = \left(\frac{n(n+1)}{2}\right)^{2}
\displaystyle\sum_{i=1}^{n}{i}^{4} = \frac{n(n+1)(2n+1)(3{n}^{2}+3{n}-1)}{30}
\displaystyle\sum_{i=1}^{n}{i}^{5} = \frac{{n}^{2}(n+1)^{2}(2{n}^{2}+2{n}-1)}{12}
\displaystyle\sum_{i=1}^{n}{i}^{6} = \frac{{n}(n+1)(2n+1)(3{n}^{4}+6{n}^{2}-3n+1)}{42}

Once the formula for a sum is known, it can be proved by induction of course. The above derivation, with a various degree of effort involved, shows the way of getting the sums directly, one after another.


Reference

  1. G. M. Boudreaux, What Do You Get When You Cross a Power Sum with an Iraqi Bank Note, Math Magazine, Vol 82, No 4 (Oct. 2009) 289-291