Prove that for {n}ge{2},{2}cdot{left(begin{matrix}{n}{2}end{matrix}right)}+{left(begin{matrix}{n}{1}end{matrix}right)}={n}^{2}

Question
Prove that for \({n}\ge{2},{2}\cdot{\left(\begin{matrix}{n}\\{2}\end{matrix}\right)}+{\left(\begin{matrix}{n}\\{1}\end{matrix}\right)}={n}^{2}\)

Answers (1)

2021-01-11
Prove that for
\({n}\ge{2},{2}\cdot{\left(\begin{matrix}{n}\\{2}\end{matrix}\right)}+{\left(\begin{matrix}{n}\\{1}\end{matrix}\right)}={n}^{2}\)
Step 1. Definitions
Definition combination
\({n}{C}{r}={\left(\begin{matrix}{n}\\{r}\end{matrix}\right)}=\frac{{{n}!}}{{{r}!{\left({n}-{r}\right)}!}}\)
with \(n!=n\times(n-1)\times\dotsb\times2\times1\).
Pascal's equation
\({\left(\begin{matrix}{n}+{1}\\{k}\end{matrix}\right)}={\left(\begin{matrix}{n}\\{k}\end{matrix}\right)}+{\left(\begin{matrix}{n}\\{k}-{1}\end{matrix}\right)}\)
Step 2. Solution
To proof: \({2}\cdot{\left(\begin{matrix}{n}\\{2}\end{matrix}\right)}+{\left(\begin{matrix}{n}\\{1}\end{matrix}\right)}={n}^{2}\) for all positive integers.
Proof by induction
Let P(n) be the statement "\({2}\cdot{\left(\begin{matrix}{n}\\{1}\end{matrix}\right)}{\left(\begin{matrix}{n}\\{1}\end{matrix}\right)}={n}^{2}\)".
Basis step n=2
\({2}\cdot{\left(\begin{matrix}{n}\\{2}\end{matrix}\right)}+{\left(\begin{matrix}{n}\\{1}\end{matrix}\right)}={2}\cdot{\left(\begin{matrix}{2}\\{2}\end{matrix}\right)}+{\left(\begin{matrix}{2}\\{1}\end{matrix}\right)}={2}\cdot{1}+{2}={4}\)
\(n^{2}=2^{2}=4\)
Thus P(2) is true.
Inductive step. Let P(k) be true.
\({2}\cdot{\left(\begin{matrix}{k}\\{2}\end{matrix}\right)}+{\left(\begin{matrix}{k}\\{1}\end{matrix}\right)}={k}^{2}\)
We need to proof that P(k+1) is true.
\({2}\cdot{\left(\begin{matrix}{k}+{1}\\{2}\end{matrix}\right)}+{\left(\begin{matrix}{k}+{1}\\{1}\end{matrix}\right)}\) \(={2}\cdot{\left({\left(\begin{matrix}{k}\\{2}\end{matrix}\right)}+{\left(\begin{matrix}{k}\\{1}\end{matrix}\right)}\right)}+{\left(\begin{matrix}{k}+{1}\\{1}\end{matrix}\right)}\ \text{Pascal's identity}\)
\(={2}\cdot{\left(\begin{matrix}{k}\\{2}\end{matrix}\right)}+{2}{\left(\begin{matrix}{k}\\{1}\end{matrix}\right)}+{\left(\begin{matrix}{k}+{1}\\{1}\end{matrix}\right)}\ \text{Distributive property}\)
\(={2}\cdot{\left(\begin{matrix}{k}\\{2}\end{matrix}\right)}+{\left(\begin{matrix}{k}\\{1}\end{matrix}\right)}+{\left(\begin{matrix}{k}\\{1}\end{matrix}\right)}+{\left(\begin{matrix}{k}+{1}\\{1}\end{matrix}\right)}\)
\(={k}^{2}+{\left(\begin{matrix}{k}\\{1}\end{matrix}\right)}+{\left(\begin{matrix}{k}+{1}\\{1}\end{matrix}\right)}\ \text{Since P(k) is true}\)
\(={k}^{2}+{k}+{k}+{1},\ {\left(\begin{matrix}{m}\\{1}\end{matrix}\right)}={m}\)
\(={k}^{2}+{2}{k}+{1}\ \text{Combine like terms}\)
\(=={\left({k}+{1}\right)}^{2},\ {\left({a}+{b}\right)}^{2}={a}^{2}+{2}{a}{b}+{b}^{2}\)
Thus P(k+1) is true.
Conclution. By the principle of mathematical induction, P(n) is true for all positive integers n.
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