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Question Number 12209 by tawa last updated on 16/Apr/17
For all n ≥ 1 ,  n ∈ Z,  prove that,   p(n) : 4 + 8 + ... + 4n = 2n(n + 1)
$$\mathrm{For}\:\mathrm{all}\:\mathrm{n}\:\geqslant\:\mathrm{1}\:,\:\:\mathrm{n}\:\in\:\mathrm{Z},\:\:\mathrm{prove}\:\mathrm{that},\: \\ $$$$\mathrm{p}\left(\mathrm{n}\right)\::\:\mathrm{4}\:+\:\mathrm{8}\:+\:…\:+\:\mathrm{4n}\:=\:\mathrm{2n}\left(\mathrm{n}\:+\:\mathrm{1}\right) \\ $$
Commented by tawa last updated on 16/Apr/17
please show me full workings sirs. God bless you all.
$$\mathrm{please}\:\mathrm{show}\:\mathrm{me}\:\mathrm{full}\:\mathrm{workings}\:\mathrm{sirs}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{all}. \\ $$
Answered by mrW1 last updated on 16/Apr/17
S=4+8+...+4n  (S/4)=1+2+...+n  (S/4)=n+...+2+1  (S/4)+(S/4)=(1+n)+...+(n+1)=n(n+1)  (S/2)=n(n+1)  ⇒S=2n(n+1)
$${S}=\mathrm{4}+\mathrm{8}+…+\mathrm{4}{n} \\ $$$$\frac{{S}}{\mathrm{4}}=\mathrm{1}+\mathrm{2}+…+{n} \\ $$$$\frac{{S}}{\mathrm{4}}={n}+…+\mathrm{2}+\mathrm{1} \\ $$$$\frac{{S}}{\mathrm{4}}+\frac{{S}}{\mathrm{4}}=\left(\mathrm{1}+{n}\right)+…+\left({n}+\mathrm{1}\right)={n}\left({n}+\mathrm{1}\right) \\ $$$$\frac{{S}}{\mathrm{2}}={n}\left({n}+\mathrm{1}\right) \\ $$$$\Rightarrow{S}=\mathrm{2}{n}\left({n}+\mathrm{1}\right) \\ $$
Commented by tawa last updated on 16/Apr/17
i appreciate sir. God bless you.
$$\mathrm{i}\:\mathrm{appreciate}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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