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Question Number 9069 by tawakalitu last updated on 16/Nov/16

Commented by RasheedSoomro last updated on 20/Nov/16

{ ((x+y=2)),((xy=4)),((S_n =x^n +y^n )) :} pS_n =S_(n+1) +qS_(n−1) p(x^n +y^n )=(x^(n+1) +y^(n+1) )+q(x^(n−1) +y^(n−1) ) Continue

$$\begin{cases}{\mathrm{x}+\mathrm{y}=\mathrm{2}}\\{\mathrm{xy}=\mathrm{4}}\\{\mathrm{S}_{\mathrm{n}} =\mathrm{x}^{\mathrm{n}} +\mathrm{y}^{\mathrm{n}} }\end{cases} \\ $$$$\mathrm{pS}_{\mathrm{n}} =\mathrm{S}_{\mathrm{n}+\mathrm{1}} +\mathrm{qS}_{\mathrm{n}−\mathrm{1}} \\ $$$$\mathrm{p}\left(\mathrm{x}^{\mathrm{n}} +\mathrm{y}^{\mathrm{n}} \right)=\left(\mathrm{x}^{\mathrm{n}+\mathrm{1}} +\mathrm{y}^{\mathrm{n}+\mathrm{1}} \right)+\mathrm{q}\left(\mathrm{x}^{\mathrm{n}−\mathrm{1}} +\mathrm{y}^{\mathrm{n}−\mathrm{1}} \right) \\ $$$$\mathrm{Continue} \\ $$

Commented by tawakalitu last updated on 20/Nov/16

Thank you sir. i will be expecting

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{will}\:\mathrm{be}\:\mathrm{expecting} \\ $$

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