Submitted by jtedds on Sun, 04/30/2017 - 16:58

I've manipulated the equation to obtain the required result in (i) and used that to find $$S_3(n) = \frac{1}{4}n^2(n+1)^2$$ and $$S_4(n) = \frac{1}{30}n(n+1)(2n+1)(3n^2+3n-1)$$ - it doesn't look like I can factorise the quadratic part of $S_4(n)$

In part (ii)

I can see that $S_k(n)$ is of degree $k+1$ since the formula shows $(n+1)^{k+1}$ and since it doesn't appear anywhere else in the equation as the other parts contain degrees of $k,k-1,k-1,\ldots\,,1$ its coefficient must be non-zero.

As the other terms also do not contain constants, the only constants come from $(n+1)^{k+1}$ and $-(n+1)$ which summing the constants eliminates it.

But I don't see how to show the sum of the coefficients is always $1$, could you give me a small pointer on that?

## Yep, you don't need to

Yep, you don't need to factorise that quadratic. Your answer is fine as it is.

Let $S_k(n)$ be the polynomial. It is then of the form $a_0 + a_1 n + a_2 n^2 + \cdots + a_{k+1}n^{k+1}$. Try evaluating this sum at $n=1$, expressing your answer in terms of the $a_i$'s?

What is the value of $S_k(1)$ from the definition of $S$?

## Since $S_k(1) =1$ for all

Since $S_k(1) =1$ for all values of k and $S_k(n) = a_0 + a_1 n + a_2 n^2 + \cdots + a_{k+1}n^{k+1}$

Then for $n =1$

$S_k(1) =a_0 + a_1 + a_2 + \cdots + a_{k+1} = 1$

Thank you!

## Yep! Remember that a very

Yep! Remember that a very typical trick to find values relating coefficients of polynomials is $f(0)$ to find constant term. $f(1)$ to find sum of coefficients. $f'(0)$ to find linear term. $f''(0)/2$ to find quadratic term, etc...