Review question

# When does the sum of this series equal $60$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8121

## Solution

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

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 $a_1 + \dotsb + a_k$ 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

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

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