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Question Number 12209 by tawa last updated on 16/Apr/17

$$\mathrm{For}\mathrm{all}\mathrm{n}⩾1,\mathrm{n}\in \mathrm{Z},\mathrm{prove}\mathrm{that},$$$$\mathrm{p}\left(\mathrm{n}\right):4+8+...+4\mathrm{n}=2\mathrm{n}(\mathrm{n}+1)$$

Commented by tawa last updated on 16/Apr/17

$$\mathrm{please}\mathrm{show}\mathrm{me}\mathrm{full}\mathrm{workings}\mathrm{sirs}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{all}.$$

Answered by mrW1 last updated on 16/Apr/17

$$S=4+8+...+4n$$$$\frac{S}{4}=1+2+...+n$$$$\frac{S}{4}=n+...+2+1$$$$\frac{S}{4}+\frac{S}{4}=(1+n)+...+(n+1)=n(n+1)$$$$\frac{S}{2}=n(n+1)$$$$\Rightarrow S=2n(n+1)$$

Commented by tawa last updated on 16/Apr/17

$$\mathrm{i}\mathrm{appreciate}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

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