Question

Prove the following by using the principle of mathematical induction for all $n\in N$
$(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})\cdots (1+\frac{1}{n})=(n+1)$

Answer 1

Step $\left(1\right):$ Assume given statement
Let the given statement be $P\left(n\right)$, i.e.
$P\left(n\right):(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3}).....(1+\frac{1}{n})=(n+1)$

Step $\left(2\right):$ Checking statement $P\left(n\right)$ for $n=1$
Put $n=1$in $P\left(n\right)$, we get
$P\left(1\right):(1+\frac{1}{1})=(1+1)$
$\Rightarrow 2=2$
Thus $P\left(n\right)$ is true for $n=1$

Step $\left(3\right):$ $P\left(n\right)$ for $n=K$
Put $n=K$in $P\left(n\right)$ and assume this is true for some natural number $K$ i.e.,
$P\left(K\right):(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})\cdots (1+\frac{1}{K})=(K+1)\cdots \left(1\right)$

Step $\left(4\right):$ Checking statement $P\left(n\right)$ for $n=K+1$
Now we shall prove that $P(K+1)$is true whenever $P\left(K\right)$ is true.
Now, we have
$(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})\cdots (1+\frac{1}{K})(1+\frac{1}{(K+1)})$
$=(K+1)(1+\frac{1}{K+1})$ (using $\left(1\right)$)
$=(K+1)\left(\frac{K+1+1}{K+1}\right)$
$=\left(\right(K+1)+1)$
Thus, $P(K+1)$ is true whenever $P\left(K\right)$ is true.
Final Answer : Therefore, by the principle of mathematical induction, statement $P\left(n\right)$ is true for all $n\in N$.