Question

Using mathematical induction, Prove
(1+x)^n$\geq$1+nx for all n$\geq$1

Answer 1

Let,

$\longrightarrow\sf{P(n):(1+x)^n\geq1+nx,\ \forall n\geq1}$

[Note: $\sf{P(n)}$ is assumed to hold true for $\sf{x\geq0.}$]

Consider $\sf{P(1).}$

$\longrightarrow\sf{(1+x)^n=1+x}$

Here $\sf{P(1)}$ is true but it only shows that $\sf{(1+x)^n=1+nx.}$

So consider $\sf{P(2).}$

$\longrightarrow\sf{(1+x)^2=1+2x+x^2}$

Since $\sf{x^2\geq0,}$

$\longrightarrow\sf{(1+x)^2\geq1+2x}$

Thus $\sf{P(n)}$ is true for $\sf{n=1}$ and $\sf{n=2.}$

So, assume $\sf{P(k)}$ is true.

$\longrightarrow\sf{P(k):(1+x)^k\geq1+kx,\ \forall k\geq1}$

Let,

$\longrightarrow\sf{(1+x)^k=1+kx+m,\ m\geq0}$

Then, consider $\sf{P(k+1).}$

$\longrightarrow\sf{P(k+1):(1+x)^{k+1}\geq1+(k+1)x,\forall k\geq1}$

Let's check whether it's true.

$\longrightarrow\sf{(1+x)^{k+1}=(1+x)(1+x)^k}$

$\longrightarrow\sf{(1+x)^{k+1}=(1+x)(1+kx+m)}$

$\longrightarrow\sf{(1+x)^{k+1}=1+kx+m+x+kx^2+mx}$

$\longrightarrow\sf{(1+x)^{k+1}=1+(k+1)x+kx^2+mx+m\quad\quad\dots(1)}$

But since $\sf{k,\ x,\ m\geq0}$

$\longrightarrow\sf{kx^2+mx+m\geq0}$

Then (1) implies,

$\longrightarrow\sf{(1+x)^{k+1}\geq1+(k+1)x}$

Therefore $\sf{P(k+1)}$ is true whenever $\sf{P(k)}$ is true.

Hence $\sf{P(n)}$ holds true $\sf{\forall n\geq1.}$