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Question Number 5722 by Rasheed Soomro last updated on 25/May/16

$$\mathrm{Prove}\mathrm{by}\mathbf{mathematical}\mathbf{induction}$$$$\mathrm{that}\mathrm{tbe}\mathrm{following}\mathrm{formula}\mathrm{is}\mathrm{correct}$$$$\mathrm{for}\mathrm{all}\mathrm{positive}\mathrm{integers}\mathrm{n}:$$$$\left(\begin{array}{c}2\\ 2\end{array}\right)+\left(\begin{array}{c}3\\ 2\end{array}\right)+\left(\begin{array}{c}4\\ 2\end{array}\right)+...+\left(\begin{array}{c}\mathrm{n}+1\\ 2\end{array}\right)=\left(\begin{array}{c}\mathrm{n}+2\\ 3\end{array}\right)$$

Commented by Yozzii last updated on 25/May/16

$$Letp\left(n\right):\underset{i=1}{\sum ^n}\left(\begin{array}{c}i+1\\ 2\end{array}\right)=\left(\begin{array}{c}n+2\\ 3\end{array}\right)(n\in \mathbb{N})$$$$$$$$Forn=1$$$$lhs=\left(\begin{array}{c}1+1\\ 2\end{array}\right)=\left(\begin{array}{c}2\\ 2\end{array}\right)=1$$$$rhs=\left(\begin{array}{c}1+2\\ 3\end{array}\right)=\left(\begin{array}{c}3\\ 3\end{array}\right)=1$$$$lhs=rhs\Rightarrow p\left(n\right)trueforn=1.$$$$$$$$Assumep\left(n\right)istrueforn=k$$$$\Rightarrow \underset{i=1}{\sum ^k}\left(\begin{array}{c}i+1\\ 2\end{array}\right)=\left(\begin{array}{c}k+2\\ 3\end{array}\right)$$$$$$$$Forn=k+1$$$$\Rightarrow \underset{i=1}{\sum ^{k+1}}\left(\begin{array}{c}i+1\\ 2\end{array}\right)=\underset{i=1}{\sum ^k}\left(\begin{array}{c}i+1\\ 2\end{array}\right)+\left(\begin{array}{c}k+2\\ 2\end{array}\right)$$$$=\left(\begin{array}{c}k+2\\ 3\end{array}\right)+\left(\begin{array}{c}k+2\\ 2\end{array}\right)$$$$=(k+2)!(\frac{1}{(k-1)!3!}+\frac{1}{k(k-1)!2!})$$$$=\frac{(k+2)!}{(k-1)!2!}(\frac{1}{3}+\frac{1}{k})$$$$=\frac{(k+2)!(k+3)}{k!3!}$$$$=\frac{(k+3)!}{(k+3-3)!3!}$$$$\underset{i=1}{\sum ^{k+1}}\left(\begin{array}{c}i+1\\ 2\end{array}\right)=\left(\begin{array}{c}k+3\\ 3\end{array}\right)$$$$\therefore p\left(k\right)\Rightarrow p(k+1).$$$$\therefore p\left(n\right)istrue\mathrm{\forall }n\in \mathbb{N}byP.M.I.$$

Commented by Rasheed Soomro last updated on 25/May/16

$$\mathcal{T}{h}^{a}nk\mathcal{S}!$$

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