Question

Using principle of mathematical induction, prove that ${41}^n-{14}^n$ is a multiple of 27.

Or
Prove by the principle of mathematical induction $n<2^n$ for all $nϵN$.

Answer 1

Let P(n) be the statement given by $P\left(n\right):{41}^n-{14}^n$ is a multiple of 27

Put n=1, we have ${41}^1-{14}^1=41-14=27$, which is multiple of 27.
So P(1) is true

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

Put n=k, we get

${41}^k-{14}^k$ is a multiple of 27.

$\Rightarrow {41}^k-{14}^k=27m\dots \left(i\right)$

We shall now prove that P(k+1) is true, whenever P(k)is true.

Now, put $n=k+1thenP(k+1)={41}^{k+1}-{41}^{k+1}$

$={41}^k,41-{14}^k.14$

$=(27m+{14}^k).41-{14}^k.14[\because {41}^k=27m+{14}^k]$

$=27m\times 41+{14}^k.41-{14}^k.14=27m\times 41+{14}^k(41-14)$

$=27m\times 41+{14}^k.27=27(41m+{14}^k)$, which is a multiple of 27

$\therefore P(k+1)$ is true.

Thus, P(k) is true $\Rightarrow P(k+1)$ is true.

Hence, by principle of mathematical induction, P(n) is $\forall nϵN$

Or

Let P(n) be the statement given by $P\left(n\right):n<2^n$

Put n=1, we get $1<2^1\Rightarrow 1<2$

$\therefore P\left(1\right)$ is true.

Assume that P(k) is true, for some natural number k, then $k<2^k$

We shall now show that P(k+1) is true, whenever P(k) is true
We have , $k<2^k$

$\Rightarrow 2k<{2.2}^k\Rightarrow k+k<2^{k+1}$

$\Rightarrow k+1\le k+k<2^{k+1}\Rightarrow k+1<2^{k+1}$

$\Rightarrow P(k+1)$ is true.

Thus, P(k) is true $\Rightarrow P(k+1)$ is true.

Hence, by principle of mathematical induction, P(n) is true, $\forall nϵN$.