`(1 + a)^n &gt;= na, n &gt;= 1 and a &gt; 0` Use mathematical induction to prove the inequality for the indicated values of n.

Friday, January 15, 2010

$\displaystyle{{\left({1}+{a}\right)}}^{{n}}\ge{n}{a},{n}\ge{1}{\quad\text{and}\quad}{a}>{0}$ Use mathematical induction to prove the inequality for the indicated values of n.

You need to use mathematical induction to prove the inequality, hence, you need to perform the following two steps, such that:

Step 1: Basis: Prove that the statement holds for n = 1

$\displaystyle{{\left({1}+{a}\right)}}^{{1}}\ge{1}\cdot{a}\Rightarrow{1}+{a}>{a}$

Step 2: Inductive step: Show that if P(k) holds, then also P(k + 1) holds.

$\displaystyle{P}{\left({k}\right)}:{{\left({1}+{a}\right)}}^{{k}}\ge{k}\cdot{a}$ holds

$\displaystyle{P}{\left({k}+{1}\right)}:{{\left({1}+{a}\right)}}^{{{k}+{1}}}\ge{\left({k}+{1}\right)}\cdot{a}$

You need to use induction hypothesis that P(k) holds, hence, you need to re-write the left side of inequality such that:

$\displaystyle{{\left({1}+{a}\right)}}^{{{k}+{1}}}={{\left({1}+{a}\right)}}^{{k}}\cdot{\left({1}+{a}\right)}\ge{k}\cdot{a}\cdot{\left({1}+{a}\right)}\ge{\left({k}+{1}\right)}\cdot{a}$

Opening the brackets yields:

$\displaystyle{k}{a}+{k}{{a}}^{{2}}\ge{k}{a}+{a}$

Notice that $\displaystyle{k}{{a}}^{{2}}>{a}$ , hence, the inequality $\displaystyle{k}{a}+{k}{{a}}^{{2}}\ge{k}{a}+{a}$ holds.

Hence, since both the basis and the inductive step hold, the statement $\displaystyle{P}{\left({n}\right)}:{{\left({1}+{a}\right)}}^{{n}}\ge{n}\cdot{a}$ holds for all indicated values of n.

Notes

Friday, January 15, 2010

$\displaystyle{{\left({1}+{a}\right)}}^{{n}}\ge{n}{a},{n}\ge{1}{\quad\text{and}\quad}{a}>{0}$ Use mathematical induction to prove the inequality for the indicated values of n.

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