Question

Prove by using the principle of mathematical induction $\forall n\in N$

$2+5+8+11+...+(3n-1)=\frac{1}{2}n(3n+1)$

Or

Using principle of mathematical induction, prove that $4^n+15n-1$is divisible by 9 for all natural numbers n.

Answer 1

Let p(n) be the statement given by

$p\left(n\right)2+5+8+11+...+(3n-1)=\frac{1}{2}n(3n+1)$

For $n=1$

$LHS:P\left(1\right)=2andRHS:p\left(1\right)=\frac{1}{2}\left(1\right)\left(3\right(1)+1)=2$

LHS=RHS

Thus p(1) is true.

Assume that p(k) is true for some natural number k.

$\therefore 2+5+8+11+...+(3k-1)=\frac{1}{2}k(3k+1)$ ...(i)

We will now show that p(k+1) is true, whenever p(k) is true.

On adding (3k+2) both sides , of Eq.(i) , we get $2+5+8+11+...+(3k-1)+(3k+2)=\frac{1}{2}k(3k+1)+(3k+2)$

$\Rightarrow 2+5+8+11+...(3k-1)+(3k+2)=\frac{3k^2+k+6k+4}{2}$

$=\frac{3k^2+3k+4k+4}{2}=\frac{(k+1)(3k+4)}{2}=\frac{1}{2}(k+1)(3k+4)$

Thus p(k+1) is true.

Hence , by principle of mathematical induction p(n) is true $\forall n\in N$.

Or

Let p(n) be the statement given by

$p\left(n\right):4+15-1=18,$ which is divisible by 9.

Thus P(1) is true.

Assume that p(k) is true for some natural number k.

$\therefore p\left(k\right):4^k+15-1$ is divisible by 9.

$\Rightarrow 4^k+15-1=9m$for some $m\in N$ ...(i)

We will now show that P(k+1) is true, whenever p(k) is true.

Now $4^{k+1}+15(k+1)-1=4^k.4+15k+15-1$

$=9m-15k+1)4+15k+14=36m-60k+4+15k+14,[usingEq.\left(1\right)]$

=$9(4m-5k+2).$which is divisible 9.

$\therefore p(k+1)$ is true, whenever p(k) is true.

Hence by principle of mathematical induction. p(n) is true, $\forall n\in N.$.