I’ve already explained in a demonstrative way how the formula for sum squared numbers arises. Not only will I only show in this article how to calculate simple series like 1+2+3+4+…., but you will also see how we enhance our findings to tackle more complicated series like 1³+2³+3³+…. These formulas are applied in many different contexts.

Simple Sum: 1+2+3+4+…

The most simple series (or in mathematical terms: partial sum) we could calculate is the sum of the first n integers. Therefore we define three series:

\displaystyle a_{n}=1+1+1+1+\dots=\sum\limits_{k=1}^{n}1=n

\displaystyle b_{n}=1+2+3+4+\dots=\sum\limits_{k=1}^{n}k

\displaystyle c_{n}=1^{2}+2^{2}+3^{2}+4^{2}+\dots=\sum\limits_{k=1}^{n}k^{2}

You may wonder why c_{n}  is needed, but often in mathematics you need to generate information from almost nothing and this is what we do here 😉 See where this approach takes us.

\displaystyle c_{n}-c_{n-1}=\sum\limits_{k=1}^{n}k^{2}-\sum\limits_{k=1}^{n-1}k^{2}=n^{2}

Now we come to the point where the magic happens 😀 The right-hand sum with upper limit n-1  can be modified. Therefore let’s add 1 to the upper limit. To make the value of the sum stay constant, we change the term in the sum from k  to k-1 . The lower limit stays at it is, because the value for k=1  is zero and, in that, doesn’t affect the sum’s value.

\displaystyle n^{2}=\sum\limits_{k=1}^{n}k^{2}-\sum\limits_{k=1}^{n}(k-1)^{2}=\sum\limits_{k=1}^{n}\big( k^{2}-(k-1)^{2}\big)

\displaystyle n^{2}=\sum\limits_{k=1}^{n}(2k-1)=2\sum\limits_{k=1}^{n}k-\sum\limits_{k=1}^{n}1

It is almost obvious how to proceed. The equation can be rearranged to make the demanded series the subject. Mission complete!

\displaystyle 2\sum\limits_{k=1}^{n}k=n^{2}+\sum\limits_{k=1}^{n}1

\displaystyle \sum\limits_{k=1}^{n}k=1+2+3+4+\dots=\frac{n^{2}+n}{2}=\frac{n(n+1)}{2}

Sum of Squares: 1²+2²+3²+4²+…

It is very important that you followed the ideas of the previous section, because – believe it or not – the shown pattern can be extended to tackle the summation of squared numbers! In addition to the series, which I already covered in the previous section, let’s define d_{n}.

\displaystyle d_{n}=1^{3}+2^{3}+3^{3}+4^{3}+\dots=\sum\limits_{k=1}^{n}k^{3}

Again we apply the magic trick involving subtraction of the highest series.

\displaystyle d_{n}-d_{n-1}=\sum\limits_{k=1}^{n}k^{3}-\sum\limits_{k=1}^{n-1}k^{3}=n^{3}

After shifting the limits like shown in the previous section, we can simplify.

\displaystyle n^{3}=\sum\limits_{k=1}^{n}\Big(k^{3}-(k-1)^{3}\Big)=\sum\limits_{k=1}^{n} 3k^{2}- \sum\limits_{k=1}^{n}3k+\sum\limits_{k=1}^{n}1=3c_{n}- 3b_{n}+a_{n}

But what’s this? We know n^{3}  on the left-hand side as well as a_{n} and b_{n} on the right-hand side! We just have to make c_{n} the subject of the formula.

\displaystyle n^{3}=3c_{n}- \frac{3n(n+1)}{2}+n

\displaystyle 3c_{n}=\frac{2n^{3} + 3n^{2}+n}{2}

\displaystyle c_{n}=1^{2}+2^{2}+3^{2}+4^{2}+\dots=\frac{n(n+1)(2n+1)}{6}

Sum of Cubes: 1³+2³+3³+4³+…

As expected, we will introduce a new series. We are looking for the sum of cube numbers, so the new series comprise numbers raised to the fourth power. I will skip the definition of e_{n} and directly show the approach:

\displaystyle e_{n}-e_{n-1}=\sum\limits_{k=1}^{n}\Big(k^{4}-(k-1)^{4}\Big)=4d_{n}-6c_{n}+4b_{n}-a_{n}=n^{4}

Making d_{n} the subject leads us to the solution.

\displaystyle d_{n}=\frac{n^{4}+6c_{n}-4b_{n}+a_{n}}{4}=\frac{n^{4}+n(n+1)(2n+1)-2n(n+1)+n}{4}

Just simplify this and receive a handy formula which allows you to add cube numbers together 😉

\displaystyle d_{n}=1^{2}+2^{2}+3^{2}+4^{2}+.\dots=\frac{n^{4}+2n^{3}+n^{2}}{4}

Sum of Higher Powers

You can extend the pattern to find formulas for sums of even higher powers. Just bear in mind that you have to introduce a series (partial sum) whose summands are raised to the power you are searching for + 1. Next, set up the difference between the elements with number n and n-1, then simplify. 😉