Sum of the first n quadratic integersTo prove thatS=∑k=1nk2=n(2n+1)(n+1)6

Question

Alannah Farmer · Accepted Answer

In mathematics, good things happen when you can evaluate something in two different ways. The telescoping series

S_{1} = \sum_{k = 0}^{n} {(k - 1)}^{3} - k^{3}

,
can be evaluated using the 'method of differences', where most of the terms are cancelled out. This tells us that

S_{1} = - n^{3}

.
However, we also know that

S_{1} = \sum_{k = 0}^{n} - 3 k^{2} + 3 k - 1

. (*)
which expresses

S_{1}

in terms of the unknown

S = \sum_{k = 1}^{n} k^{2}

.
Because we already know that

S_{1} = - n^{3}

using the method of differences, we can rearrange (*) to make S the subject of the equation and solve.
The same principle applies when computing series such as

\sum_{k = 1}^{n} k^{3}

: because we can evaluate

\sum_{k = 1}^{n} {(k - 1)}^{4} - k^{4}

in two different ways, this gives us enough information to find

\sum_{k = 1}^{n} k^{3}

.

horieblersee275 · Accepted Answer

This is a more general answer than what you are actually asking.
The reason is for me as follows: you only know the sums for

e l l < e l l_{0}

, and you want to calculate

\sum_{k = 1}^{n} k^{e l l_{0}}

The terms in

{(k + 1)}^{e l l_{0} + 1} - k^{e l l_{0} + 1}

are all of the form

α_{e l l} k^{e l l}

with

e l l \leq e l l_{0}

so you know how to calculate each

α_{e l l} \sum_{k = 1}^{n} k^{e l l} \forall e l l < e l l_{0}

Except

α e l l_{0} \sum_{k = 1}^{n} k^{e l l_{0}}

Moreover,

\sum_{k = 1}^{n} ({(k + 1)}^{e l l_{0} + 1} - k^{e l l_{0} + 1}

is a telescoping sum so it is really easy to calculate.
You can see that

\sum_{k = 1}^{n} ({(k + 1)}^{e l l_{0} + 1} - k^{e l l_{0} + 1}) = {(n + 1)}^{e l l_{0} + 1} - 1

Therefore it gives you enough information to determine,

\sum_{k = 1}^{n} k^{e l l_{0}}

Sum of the first n quadratic integers To prove

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