Mathematical Induction

 

Prove by mathematical induction \[ (1+x)^n \ge 1+nx \quad \text{for} \quad x \gt -1 \]

Prove the statement is true for a particular n. Here n = 1 is an obvious choice.

To show this is true for n=1, observe that the left hand side = 1 + x and the right hand side = 1 + x.
So the statement is true for n=1.

Assume the statement is true for n = k and show it is true for n = k + 1.

To show the statement is true for n = k+1, set \[ \begin{align} (1+x)^{k+1} &= (1+x)^k(1+x) \\ \end{align} \ \]

Use the inductive hypothesis to simplify the right hand side.

\[ \begin{align} (1+x)^{k+1} &= (1+x)^k(1+x) \quad \\ &\ge (1+kx)(1+x) \quad (\text{ind. hyp. and} \quad 1 + x > 0) \quad \\ &= 1 + (k+1)x + kx^2 \quad \\ &\ge 1 + (k+1)x \quad \text{(because} \quad kx^2 \ge 0) \quad \\ \end{align} \ \] So the statement is true for n = k + 1.

By the principle of mathematical induction,

The statement is true for \( n \ge 1 \).

You will gain most benefit if you work through each problem first and then come back and check your work against the solution.

 

y

Now You Try:

Prove by mathematical induction \[ \frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} \lt 2 - \frac{2}{n} \] for all sufficiently large n > 0.
State the smallest value of n for which this is true.