I was solving A-level Further Mathematics paper and I didn't quite understand how to solve the question.
Question:
I know the formula $\sum _{r=1}^n(u_{r})=f\left(1\right)-f\left(n+1\right)$ but I can't seem to figure out how to use it.
This was the answer provided. It isn't very clear on how to solve the problem! Answer:
Could you provide me an explanation on how to approach this type of problem. Thanks.
With the given identity
$(n+\frac12)^6-(n-\frac12)^6=6n^5+5n^3+\frac38n\Leftrightarrow n^5=\frac16\left((n-\frac12)^6-(n+\frac12)^6+5n^3+\frac38n\right)$
We get:
$\sum_{n=1}^N n^5=\sum_{n=1}^N \frac16\left((n-\frac12)^6-(n+\frac12)^6+5n^3+\frac38n\right)$
With basic sum manipulation we derive:
$=\frac16\sum_{n=1}^N (n-\frac12)^6-(n+\frac12)^6+\frac56\sum_{n=1}^N n^3+\frac1{16}\sum_{n=1}^N n$
Note that the first sum is telescoping. The last sum, is the well known Gaussian sum $\sum_{n=1}^N n=\frac{N(N+1)}{2}$
Depending on what you know, it is also known, that
$\sum_{n=1}^N n^3=\frac{N^2(N+1)^2}{4}$
If you know this result, we are kinda done and you just have to calculate the telescoping sum, which is not difficult and substitute the other sums with known formulas. If you do not know these formulas, you might proof them first.
This can be down with induction fairly easy, but would a little bit trick the question, in my opinion.
You can also use the same trick to calculate $\sum_{n=1}^N n^3$, by writing it as a telesoping sum. Then you just need to know $\sum_{n=1}^N n^2$ and $\sum_{n=1}^N n$.
There are many ways, it depends on your background.
Just sum the given identity for all $n=1,2,...,N$. LHS will telescope and on the RHS you will obtain six times your desired sum plus five times the sum of cubes (which is known) and $\frac{3}{8}$ of the sum of the first powers which is also known. So you can solve for the sum you are looking for.
