## Insight Things

### A scientific blog revealing the hidden links which shape our world

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 😉

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

### 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. 😉