Question

Using mathematical induction, prove that $(1 + \frac{3}{1})(1 + \frac{5}{4})(1 + \frac{7}{9})......(1 + \frac{2n + 1}{n^{2}}) = (n + 1)^{2}$ for all n ∈ N

Answer 1

Mathematical Induction

We are given a statement which we are supposed to prove by the Principle of Mathematical Induction.

Let the statement be P(n).

$\displaystyle\sf P(n): \left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)\dots\left(1+\frac{2n+1}{n^2}\right)=(n+1)^2$

Checking P(1):

$\begin{array}{l|l}\mathbb{LHS}&\mathbb{RHS}\\\cline{1-2}&\\\\\ = 1+\dfrac{3}{1}& = (1+1)^2\\\\ = 4 & = 4 \end{array}$

Thus, P(1) is true.

Suppose P(k) is true.

$\displaystyle\sf \implies\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)\dots\left(1+\frac{2k+1}{k^2}\right)=(k+1)^2 \quad \textsf{-----(1)}$

We now need to check for P(k+1).

To Prove:

$\displaystyle\sf\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)\dots\left(1+\frac{2(k+1)+1}{(k+1)^2}\right)=((k+1)+1)^2$

Consider the LHS:

$\displaystyle\sf\mathbb{LHS}\\\\\\ =\underbrace{\sf\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)\dots\left(1+\frac{2k+1}{k^2}\right)}_\textsf{Use Result (1)}\left(1+\frac{2(k+1)+1}{(k+1)^2}\right)\\\\\\ = \cancel{(k+1)^2} \left(\frac{(k+1)^2+2(k+1)+1)}{\cancel{(k+1)^2}}\right)\\\\\\ \textsf{Use the identity: }a^2+2ab+b^2=(a+b)^2\\\\\\ = ((k+1)+1)^2\\\\\\ = \mathbb{RHS}$

Thus, P(k+1) is true provided P(k) is true.

Now, P(1) is true and P(k) is true $\implies$ P(k+1) is true.

Hence, P(n) is true for all $\sf n\in\mathbb{N}$.

Hence Proved, by the Principle of Mathematical Induction.