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Prove-by-mathematical-induction-that-tbe-following-formula-is-correct-for-all-positive-integers-n-2-2-3-2-4-2-n-1-2-n-2-3-

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Question Number 5722 by Rasheed Soomro last updated on 25/May/16
Prove by mathematical induction that tbe following formula is correct for all positive integers n: ((2),(2) ) + ((3),(2) ) + ((4),(2) ) +...+ (((n+1)),(( 2)) ) = (((n+2)),(( 3)) )
$$\mathrm{Prove}\:\mathrm{by}\:\boldsymbol{\mathrm{mathematical}}\:\boldsymbol{\mathrm{induction}} \\ $$$$\mathrm{that}\:\mathrm{tbe}\:\mathrm{following}\:\mathrm{formula}\:\mathrm{is}\:\mathrm{correct} \\ $$$$\mathrm{for}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{n}: \\ $$$$\begin{pmatrix}{\mathrm{2}}\\{\mathrm{2}}\end{pmatrix}\:+\begin{pmatrix}{\mathrm{3}}\\{\mathrm{2}}\end{pmatrix}\:+\begin{pmatrix}{\mathrm{4}}\\{\mathrm{2}}\end{pmatrix}\:+…+\begin{pmatrix}{\mathrm{n}+\mathrm{1}}\\{\:\:\:\mathrm{2}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{n}+\mathrm{2}}\\{\:\:\:\mathrm{3}}\end{pmatrix} \\ $$
Commented by Yozzii last updated on 25/May/16
Let p(n): Σ_(i=1) ^n (((i+1)),(2) )= (((n+2)),(3) ) (n∈N) For n=1 lhs= (((1+1)),(2) )= ((2),(2) )=1 rhs= (((1+2)),(3) )= ((3),(3) )=1 lhs=rhs⇒p(n) true for n=1. Assume p(n) is true for n=k ⇒Σ_(i=1) ^k (((i+1)),(2) )= (((k+2)),(3) ) For n=k+1 ⇒Σ_(i=1) ^(k+1) (((i+1)),(2) )=Σ_(i=1) ^k (((i+1)),(2) )+ (((k+2)),(2) ) = (((k+2)),(3) )+ (((k+2)),(2) ) =(k+2)!((1/((k−1)!3!))+(1/(k(k−1)!2!))) =(((k+2)!)/((k−1)!2!))((1/3)+(1/k)) =(((k+2)!(k+3))/(k!3!)) =(((k+3)!)/((k+3−3)!3!)) Σ_(i=1) ^(k+1) (((i+1)),(2) )= (((k+3)),(3) ) ∴ p(k)⇒p(k+1). ∴p(n) is true ∀n∈N by P.M.I.
$${Let}\:{p}\left({n}\right):\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\begin{pmatrix}{{i}+\mathrm{1}}\\{\mathrm{2}}\end{pmatrix}=\begin{pmatrix}{{n}+\mathrm{2}}\\{\mathrm{3}}\end{pmatrix}\:\:\left({n}\in\mathbb{N}\right) \\ $$$$ \\ $$$${For}\:{n}=\mathrm{1} \\ $$$${lhs}=\begin{pmatrix}{\mathrm{1}+\mathrm{1}}\\{\mathrm{2}}\end{pmatrix}=\begin{pmatrix}{\mathrm{2}}\\{\mathrm{2}}\end{pmatrix}=\mathrm{1} \\ $$$${rhs}=\begin{pmatrix}{\mathrm{1}+\mathrm{2}}\\{\mathrm{3}}\end{pmatrix}=\begin{pmatrix}{\mathrm{3}}\\{\mathrm{3}}\end{pmatrix}=\mathrm{1} \\ $$$${lhs}={rhs}\Rightarrow{p}\left({n}\right)\:{true}\:{for}\:{n}=\mathrm{1}. \\ $$$$ \\ $$$${Assume}\:{p}\left({n}\right)\:{is}\:{true}\:{for}\:{n}={k} \\ $$$$\Rightarrow\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}\begin{pmatrix}{{i}+\mathrm{1}}\\{\mathrm{2}}\end{pmatrix}=\begin{pmatrix}{{k}+\mathrm{2}}\\{\mathrm{3}}\end{pmatrix}\: \\ $$$$ \\ $$$${For}\:{n}={k}+\mathrm{1} \\ $$$$\Rightarrow\underset{{i}=\mathrm{1}} {\overset{{k}+\mathrm{1}} {\sum}}\begin{pmatrix}{{i}+\mathrm{1}}\\{\mathrm{2}}\end{pmatrix}=\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}\begin{pmatrix}{{i}+\mathrm{1}}\\{\mathrm{2}}\end{pmatrix}+\begin{pmatrix}{{k}+\mathrm{2}}\\{\mathrm{2}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\begin{pmatrix}{{k}+\mathrm{2}}\\{\mathrm{3}}\end{pmatrix}+\begin{pmatrix}{{k}+\mathrm{2}}\\{\mathrm{2}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left({k}+\mathrm{2}\right)!\left(\frac{\mathrm{1}}{\left({k}−\mathrm{1}\right)!\mathrm{3}!}+\frac{\mathrm{1}}{{k}\left({k}−\mathrm{1}\right)!\mathrm{2}!}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left({k}+\mathrm{2}\right)!}{\left({k}−\mathrm{1}\right)!\mathrm{2}!}\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{{k}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left({k}+\mathrm{2}\right)!\left({k}+\mathrm{3}\right)}{{k}!\mathrm{3}!} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left({k}+\mathrm{3}\right)!}{\left({k}+\mathrm{3}−\mathrm{3}\right)!\mathrm{3}!} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{k}+\mathrm{1}} {\sum}}\begin{pmatrix}{{i}+\mathrm{1}}\\{\mathrm{2}}\end{pmatrix}=\begin{pmatrix}{{k}+\mathrm{3}}\\{\mathrm{3}}\end{pmatrix} \\ $$$$\therefore\:{p}\left({k}\right)\Rightarrow{p}\left({k}+\mathrm{1}\right). \\ $$$$\therefore{p}\left({n}\right)\:{is}\:{true}\:\forall{n}\in\mathbb{N}\:{by}\:{P}.{M}.{I}. \\ $$
Commented by Rasheed Soomro last updated on 25/May/16
Th^a nkS!
$$\mathcal{T}{h}^{{a}} {nk}\mathcal{S}! \\ $$

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