Question

Prove that by mathematical induction $2^{3n}-1$ is divisible by $7$ for all natural numbers.

Answer 1

$P\left(n\right):2^{3n}-1$

If It's divisible by $7$ then,

Base case:

$n=1$ then,

$P\left(n\right):2^{3n}-1=2^{3\times 1}-1=7$

If $n=k$ is true for some natural numbers, then It should also be true for $(k+1)$,

For $n=k$,

$P\left(k\right):2^{3k}-1=7d$

$P\left(k\right):2^{3k}=7d+1$ .....(1)

Where $d=1,2,3,\mathrm{4.......}soon$

Now we check for $n=k+1$,

$P(k+1):2^{3(k+1)}-1=7d$

$P(k+1):2^{3k+3}=7d+1$

(Since $a^{xy}=a^x{a}^y$)

$P(k+1):2^{3k}{2}^3=7d+1$

Substituting value from equation (1),

$P(k+1):(7d+1)\times 8=7d+1$

$56d+8=7d+1$

$7\times (8d+1)=7d$

So, it is true for both $kand(k+1)$.

Hence, from the principle of mathematical induction,

$2^{3n}-1$ is divisible by $7$ for all natural numbers.