When does the sum of this series equal $60$?

When does the sum of this series equal $60$?

Question

Which positive integer $n$ satisfies the equation $\frac{3}{n^3}+\frac{4}{n^3}+\frac{5}{n^3}+\dots b +\frac{n^3-5}{n^3}+\frac{n^3-4}{n^3}+\frac{n^3-3}{n^3}=60?$

Suggestion

What type of sequence do we have on the left hand side of the equation? How many terms are there in the sequence?

Solution

This is an arithmetic progression with common difference $\dfrac{1}{n^3}$. There are $n^3-5$ terms.

The formula for the sum of $k$ terms of an arithmetic progression $a_1 + \dots b + a_k$ is given by $\dfrac{k}{2}\left(a_1 + a_k\right).$

Alternatively, if the first term is $a$, and the common difference is $d$, and if there are $n$ terms, then the sum of the first $n$ terms is $S_n = \dfrac{n}{2}(2a + (n-1)d)$.

Their sum is therefore $60=\frac{n^3-5}{2}\left(\frac{3}{n^3}+\frac{n^3-3}{n^3}\right)=\frac{n^3-5}{2}\left(\frac{n^3}{n^3}\right)=\frac{n^3-5}{2},$ so we can rearrange to get $n^3-5=120\implies n^3=125\implies n=5.$

Alternatively, we could have written the sum as

$$\frac{3}{n^3}+\frac{4}{n^3}+\frac{5}{n^3}+\dots b +\frac{n^3-5}{n^3}+\frac{n^3-4}{n^3}+\frac{n^3-3}{n^3}=\frac{3+4+\dots b+(n^3-3)}{n^3}$$

Now we can treat the numerator as an arithmetic series giving $\frac{\frac{(n^3-3)(n^3-2)}{2}-1-2}{n^3} =\frac{n^6-5 n^3+6-6}{2 n^3} =\frac{n^3-5}{2} = 60, n = 5.$