Question

Use generalized induction as was done in Example 13 to show that if ${\mathbf{a}}_{\mathit{m}\mathbf{,}\mathit{n}}$is defined recursively by localid="1668616926905" ${\mathbf{a}}_{\mathbf{1}\mathbf{,}\mathbf{1}}\mathbf{=}\mathbf{5}$and ${\mathbf{a}}_{\mathbf{m}\mathbf{,}\mathbf{n}}\mathbf{=}\{\begin{array}{l}{a}_{m-1,n}+2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}n=1\text{and}m>1\\ a_{m,n-1}+2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}n>1\end{array}$

then${\mathbf{a}}_{\mathbf{m}\mathbf{,}\mathbf{n}}\mathbf{=}\mathbf{2}\left(m+n\right)\mathbf{+}\mathbf{1}\mathbf{}\mathbf{for}\mathbf{}\mathbf{all}\mathbf{}\mathbf{(}\mathbf{m}\mathbf{,}\mathbf{n}\mathbf{)}\mathbf{\in }{\mathbf{Z}}^{\mathbf{+}}\mathbf{\times }{\mathbf{Z}}^{\mathbf{+}}$

Short answer

It has been proved.

Step-by-step solution

Introduction

Mathematical Induction is a mathematical technique which is used to prove a statement, a formula or a theorem is true for every natural number. Step 1(Base step) − It proves that a statement is true for the initial value.

Basis step

This is true because

$\begin{array}{r}{a}_{1.1}=5\\ =2(1+1)+1\end{array}$

Inductive step

Assume that $a_{m^{\mathrm{\prime }},n^{\mathrm{\prime }}}=2\left(m^{\mathrm{\prime }}+n^{\mathrm{\prime }}\right)+1$whenever

(m' + n') < (m , n)

If n = 1 then

$\begin{array}{r}{\mathrm{a}}_{\mathrm{m},\mathrm{n}}={\mathrm{a}}_{\mathrm{m}-1,\mathrm{n}}+2\\ =2(\mathrm{m}-1+\mathrm{n})+1+2\\ =2\mathrm{m}-2+2\mathrm{n}+3\\ =2\mathrm{m}+2\mathrm{n}+1\\ =2(\mathrm{m}+\mathrm{n})+1\end{array}$

If n > 1 then,

$\begin{array}{r}{\mathrm{a}}_{\mathrm{m},\mathrm{n}}={\mathrm{a}}_{\mathrm{m},\mathrm{n}-1}+2\\ =2(\mathrm{m}+\mathrm{n}-1)+1+2\\ =2\mathrm{m}+2\mathrm{n}-2+1+2\\ =2\mathrm{m}+2\mathrm{n}+1\\ =2(\mathrm{m}+\mathrm{n})+1\end{array}$

Hence, proved