Question Number 9069 by tawakalitu last updated on 16/Nov/16
Commented by RasheedSoomro last updated on 20/Nov/16
$$\begin{cases}{\mathrm{x}+\mathrm{y}=\mathrm{2}}\\{\mathrm{xy}=\mathrm{4}}\\{\mathrm{S}_{\mathrm{n}} =\mathrm{x}^{\mathrm{n}} +\mathrm{y}^{\mathrm{n}} }\end{cases} \\ $$$$\mathrm{pS}_{\mathrm{n}} =\mathrm{S}_{\mathrm{n}+\mathrm{1}} +\mathrm{qS}_{\mathrm{n}−\mathrm{1}} \\ $$$$\mathrm{p}\left(\mathrm{x}^{\mathrm{n}} +\mathrm{y}^{\mathrm{n}} \right)=\left(\mathrm{x}^{\mathrm{n}+\mathrm{1}} +\mathrm{y}^{\mathrm{n}+\mathrm{1}} \right)+\mathrm{q}\left(\mathrm{x}^{\mathrm{n}−\mathrm{1}} +\mathrm{y}^{\mathrm{n}−\mathrm{1}} \right) \\ $$$$\mathrm{Continue} \\ $$
Commented by tawakalitu last updated on 20/Nov/16
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{will}\:\mathrm{be}\:\mathrm{expecting} \\ $$