Question

Please solve it... Question 6
Using the principle of mathematical induction prove that the following is true for all n natural numbers

Answer 1

So we have,

$P(n):\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)\dots\left(1-\dfrac{1}{(n+1)^2}\right)=\dfrac{n+2}{2(n+1)}$

Consider $P(1).$

$\longrightarrow LHS=1-\dfrac{1}{2^2}=\dfrac{3}{4}$

$\longrightarrow RHS=\dfrac{1+2}{2(1+1)}=\dfrac{3}{4}$

Hence $P(1)$ is true. So assume $P(k)$ is true.

$P(k):\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)\dots\left(1-\dfrac{1}{(k+1)^2}\right)=\dfrac{k+2}{2(k+1)}$

Consider $P(k+1).$

$P(k+1):\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)\dots\left(1-\dfrac{1}{(k+2)^2}\right)=\dfrac{k+3}{2(k+2)}$

Let's check whether it's true.

$\longrightarrow LHS=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)\dots\left(1-\dfrac{1}{(k+1)^2}\right)\left(1-\dfrac{1}{(k+2)^2}\right)$

From $P(k),$

$\longrightarrow LHS=\dfrac{k+2}{2(k+1)}\left(1-\dfrac{1}{(k+2)^2}\right)$

$\longrightarrow LHS=\dfrac{k+2}{2(k+1)}\cdot\dfrac{(k+2)^2-1}{(k+2)^2}$

$\longrightarrow LHS=\dfrac{(k+2)^2-1^2}{2(k+1)(k+2)}$

$\longrightarrow LHS=\dfrac{(k+2+1)(k+2-1)}{2(k+1)(k+2)}$

$\longrightarrow LHS=\dfrac{(k+3)(k+1)}{2(k+1)(k+2)}$

$\longrightarrow LHS=\dfrac{k+3}{2(k+2)}$

$\longrightarrow LHS=RHS$

Hence $P(k+1)$ is true whenever $P(k)$ is true.

Therefore $P(n)$ is true $\forall\,n\in\mathbb{N}.$