Embibe Experts Solutions for Chapter: Mathematical Induction, Exercise 1: Level 1

A student was asked to prove a statement by induction. He proved

(i) $P\left(5\right)$ is true and

(ii) Truth of $P\left(n\right)\Rightarrow$truth of $p\left(n+1\right), n\in N$

On the basis of this, he could conclude that $p\left(n\right)$ is true for

If $P\left(n\right):1+3+5+\dots +\left(2n-1\right)=n^2$, then $P\left(n\right)$ is

Let $P\left(n\right)$ denotes the statement that $n^2+n, n\in N$ is odd. It is seen that $P\left(n\right)\Rightarrow P\left(n+1\right),$ then $P\left(n\right)$ is true for all

Using mathematical induction, the numbers $a_n$'s are defined by $a_0=1, a_{n+1}= 3n^2+n+a_n, \left(n\ge 0\right)$, then $a_n=$

If $a_n=\sqrt{7+\sqrt{7+\sqrt{7+\mathrm{......}}}}$  having $n$ radical signs, then by methods of mathematical induction which of the following is true?

If $P\left(n\right):1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\dots +\frac{1}{n^2}<2-\frac{1}{n}$ is true and $n\in N,$ then

For all $n\in N, 2·4^{2n+1}+3^{3n+1}$ is divisible by

If $P\left(n\right):2+4+6+\mathrm{...}+\left(2n\right), n\in N$ then $P\left(k\right)=k\left(k+1\right)+2$ implies $P\left(k+1\right)=\left(k+1\right)\left(k+2\right)+2$ is true for all $k\in N$. So, the statement $P\left(n\right)=n\left(n+1\right)+2$ is true for