Chapter 12, Problem 14CT

To determine

To find: The statement is true or not for all natural numbers.

Answer to Problem 14CT

The statement is true for all natural numbers.

Explanation of Solution

Given information:

The given series is $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(1+\frac{1}{n}\right)=n+1$ .

Calculation:

Calculate the series,

The series is $1+\frac{1}{1}=2$ .

Substitute the value of $1$ for $n$ in $n+1$ to check the statement is true for $n=1$ .

  $1+1=2$

Therefore, the statement is true for $n=1$ .

Assume that the statement is true for $n=k$ .

Therefore,

  $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(1+\frac{1}{k}\right)=n+k$

The sum for $k+1$ terms and check the equal $k+2$ .

  $\begin{array}{c}\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(1+\frac{1}{k}\right)\left(1+\frac{1}{k+1}\right)=\left(k+1\right)\left(\frac{k+1+1}{k+1}\right)\\ =\left(k+1\right)\left(\frac{k+2}{k+1}\right)\\ =k+2\end{array}$

The statement is true for $k+1$ .

Both the conditions of the principle of Mathematical Induction are satisfied.

Therefore, the statement is true for all natural numbers.