Problem source

The equation 
\[
x^{n}-qx^{n-1}+r=0,
\]
 where $n\geqslant5$ and $q$ and $r$ are real constants, has roots $\alpha_{1},\alpha_{2},\ldots,\alpha_{n}.$ The sum of the products of $m$ distinct roots is denoted by $\Sigma_{m}$ (so that, for example, $\Sigma_{3}=\sum\alpha_{i}\alpha_{j}\alpha_{k}$ where the sum runs over the values of $i,j$ and $k$ with $n\geqslant i>j>k\geqslant1$).
The sum of $m$th powers of the roots is denoted by $S_{m}$ (so that, for example, $S_{3}=\sum\limits_{i=1}^{n}\alpha_{i}^{3}$). 
Prove that $S_{p}=q^{p}$ for $1\leqslant p\leqslant n-1.$ 

\textit{You may assume that for any $n$th degree equation and $1\leqslant p\leqslant n$}
\[
S_{p}-S_{p-1}\Sigma_{1}+S_{p-2}\Sigma_{2}-\cdots+(-1)^{p-1}S_{1}\Sigma_{p-1}+(-1)^{p}p\Sigma_{p}=0.]
\]
Find expressions for $S_{n},$ $S_{n+1}$ and $S_{n+2}$ in terms
of $q,r$ and $n$. Suggest an expression for $S_{n+m},$ where $m < n$,
and prove its validity by induction.

Solution source

Claim: $S_p = q^p$ for $1 \leq p \leq n-1$

Proof: When $p = 1$, $S_p = \Sigma_1 = q$ as expected.

Note that $\Sigma_i = 0$ for $i = 2, \cdots, n-1$.

Using $S_p = S_{p-1}\Sigma_{1}-S_{p-2}\Sigma_{2}+\cdots+(-1)^{p-1+1}S_{1}\Sigma_{p-1}+(-1)^{p+1}p\Sigma_{p}$, we can see that

$S_p = qS_{p-q}$ when $1 \leq p \leq n-1$, ie $S_p = q^p$.

Note that 
\begin{align*}
S_n &= \sum \alpha_i^n \\
&= q\sum \alpha_i^{n-1} - \sum r \\
&= qS_{n-1} - nr \\
&= q^n - nr \\
\\
S_{n+1} &=  \sum \alpha_i^{n+1} \\
&= q \sum \alpha_i^{n} - r \sum \alpha_i \\
&= q^{n+1} - rq \\
\\
S_{n+2} &= \sum \alpha_i^{n+2} \\
&= q \sum \alpha_i^{n+1} - r \sum \alpha_i^2 \\
&= q^{n+2} - rq^2 \\
\end{align*}

Claim: $S_{n+m} = q^{n+m} - rq^{m}$

Proof: The obvious