Question

Prove by the principle of mathematical induction that $(2n+7)<(n+3{)}^2$ for all natural numbers n.
Or
Prove by the principle of mathematical induction that n (n + 1) (2n + 1) is divisible by 6 for all $nϵN.$

Answer 1

Let P(n) be the statement given by
P(n) : $(2n+7)<(n+3{)}^2$
For n $=1,P\left(1\right):2+7<(1+3{)}^2\Rightarrow 9<16$
$\therefore P\left(1\right)$ is true.
Assume that P(k) is true, for some natural number k.
Then, $(2k+7)<(k+3{)}^2$
Now, we will show that P(k + 1) is true, whenever P(k) is true. For tis, we have to show that
$2(k+1)+7<(k+1+3{)}^2$
Now consider P(k) is true.
$\Rightarrow 2k+7<(k+3{)}^2\Rightarrow 2k+7+2<(k+3{)}^2+2\left[\text{adding 2 on both sides}\right]\Rightarrow 2(k+1)+7<k^2+6k+9+2\Rightarrow 2(k+1)+7<k^2+6k+11<k^2+8k+16\Rightarrow (2k+1)+7<k^2+8k+16\Rightarrow (2k+1)+7<(k+1+3{)}^2\Rightarrow P(k+1)\text{is true}$
Hence, by principle of mathematical induction, P(n) is true $\forall nϵN$

OR
Let P (n) be the statement$n(n+1)(2n+1)\text{is divisible by 6.}$
$\Rightarrow k(k+1)(2k+1)=6m\Rightarrow k(2k^2+3k+1)=6m\Rightarrow 2k^3+3k^2+k=6m$
Now, we will show that P(k + 1) is true whenever P (k) is true. For this, we have to show that (k + 1) (k + 2) 2k +3 is divisible
$\text{we have},(k+1)(k+2)(2k+3)=(k+1)(2k^2+7k+6)=2k^3+7k^2+6k+2k^2+7k+6=2k^3+9k^2+13k+6=(6m-3k^2-k)+9k^2+13k+6=6m+6k^2+12k+6=6(m+k^2+2k+1)\text{which is divisible by 6}\Rightarrow P(k+1)\text{is true}$