A Level Further Maths: Series Questions

Just can't seem to work out the answers to these questions, hope someone here can help

1) a. By considering ∑((𝑟 + 1)^5 − 𝑟^5), show that
n
∑ 𝑟^4 = 1/30𝑛(𝑛 + 1)(2𝑛 + 1)(3𝑛^2 + 3𝑛 − 1)
r=1

1) b. Hence find the value of 50^4 + 51^4 + ⋯+ 80^4

Well, tackle them one by one. For the first one, you can use Method of Differences to state what

$\displaystyle \sum_{r=1}^n (r+1)^5 - r^5$

is in terms of $n$. Let's call this result $f(n)$ because I cannot be bothered to work it out and I'm leaving it to you intentionally.

Once you have that, a realisation is required to notice that $(r+1)^5 - r^5 = 5r^4 + 10r^3 + 10r^2 + 5r + 1$

Therefore;

$\displaystyle \sum_{r=1}^n (r+1)^5 - r^5 = f(n) = \sum_{r=1}^n 5r^4 + 10r^3 + 10r^2 + 5r + 1$

And you can treat the RHS as $\displaystyle 5\sum_{r=1}^n r^4 + 10\sum_{r=1}^n r^3 + 10\sum_{r=1}^nr^2 + 5\sum_{r=1}^n r + \sum_{r=1}^n 1$.

You can determine all of these sums except the $r^4$ one. Once you do that, just rearrange the equality

$f(n) =\displaystyle 5\sum_{r=1}^n r^4 + 10\sum_{r=1}^n r^3 + 10\sum_{r=1}^nr^2 + 5\sum_{r=1}^n r + \sum_{r=1}^n 1$

for $\displaystyle \sum_{r=1}^n r^4$

Any clue how to approach the questions 3 a b and c

Have you ever covered Partial Fraction Decomposition ?

not at all

Then you can't do it.

Wow my teacher is very clever 🙄

If you have covered Method of Difference (or telescoping sums) you can do parts (b) and (c) by just taking the result of part (a) at face value.

yeah we have covered method of difference. surely part c is just infinity because u can't add numbers to it ect

Not quite. If you work out the answer to part (b) correctly then you would notice that $f(n)$ is a rational fraction with a linear term in the numerator, and a quadratic term in the denominator.

The upper limit of $\infty$ can be rewritten as a limit; i.e. $\displaystyle \sum_{r=1}^{\infty} = \lim_{n\to\infty} \sum_{r=1}^n$.

Just like improper integrals are written in terms of limits (if you have covered those). And clearly you will need to be taking the limit of $f(n)$. Key point here is that $f(n) \to 0$ in this limit precisely because of what type of function it is.

So, f(n) disappears and the sum at infinity is just a finite value, whatever is leftover.