Question

If ${7^{2n}} + {2^{3n - 3}}{.3^{n - 1}}$ is divisible by 25 for any natural number $n \geqslant 1$. Prove that by mathematics.

Answer 1

Hint: To prove that the given equation is divisible by 25. First, we had to assume the value of n = 1, and after that assume that the given equation is also divisible by 25 when n = k. Then on putting the value of n = k+1, if the given equation is divisible by 25 then the equation will be divisible by 25 for all natural numbers.

Complete step-by-step answer:
Let us prove that ${7^{2n}} + {2^{3n - 3}}{.3^{n - 1}}$ is always divisible by 25 for all natural number n by using mathematical induction.
Now as we know that to prove any equation by mathematical induction we have to first assume the value of the variable is 1 and check whether the condition satisfies the variable equal to 1 and after that we assume that the given condition also satisfies at variable = k. And then if the given equation also satisfies the given condition at the value of variable = k+1 then the result will be true.
So, here the variable in the given equation is n.
So, let $P\left( n \right) = {7^{2n}} + {2^{3n - 3}}{.3^{n - 1}}$ (1)
So, now putting the value of n equal to 1 in equation 1.
$\Rightarrow P\left( 1 \right) = {7^{2 \times 1}} + {2^{3 \times 1 - 3}}{.3^{1 - 1}} = 49 + 1 = 50 = 2 \times 25$
So, the $P\left( 1 \right)$ is divisible by 25. Hence, $P\left( 1 \right)$ is true.
So, let $P\left( k \right)$ will also be true.
So, $P\left( k \right) = {7^{2k}} + {2^{3k - 3}}{.3^{k - 1}}$ is divisible by 25.
If the above equation is divisible by 25 then the value of the above equation must be equal to 25m where m is a natural number.
$\Rightarrow {7^{2k}} + {2^{3k - 3}}{.3^{k - 1}} = 25m$
$\Rightarrow \left( {{2^{3k - 3}}} \right).\left( {{3^{k - 1}}} \right) = 25m - {7^{2k}}$ (2)
Now to show that $P\left( {k + 1} \right)$ is also true. Let us put the value of n = k+1 in equation 1.
$\Rightarrow P\left( {k + 1} \right) = {7^{2\left( {k + 1} \right)}} + {2^{3\left( {k + 1} \right) - 3}}{.3^{\left( {k + 1} \right) - 1}}$
$\Rightarrow P\left( {k + 1} \right) = {7^2}{.7^{2k}} + {2^3}{.3^1}{.2^{3k - 3}}{.3^{k - 1}}$
$\Rightarrow P\left( {k + 1} \right) = 49 \times {7^{2k}} + 24 \times {2^{3k - 3}}{.3^{k - 1}}$
Now putting the value of ${2^{3k - 3}}{.3^{k - 1}}$ in the above equation from equation 2.
$\Rightarrow P\left( {k + 1} \right) = 49 \times {7^{2k}} + 24\left( {25m - {7^{2k}}} \right)$
Solving above equation.
$\Rightarrow P\left( {k + 1} \right) = \left( {49 - 24} \right){7^{2k}} + 25\left( {24m} \right)$
$\Rightarrow P\left( {k + 1} \right) = 25\left( {{7^{2k}} + 24m} \right)$
Now as we can see that the above equation is also divisible by 25.
So, $P\left( {k + 1} \right)$ is also true whenever $P\left( k \right)$ is true.
Hence, by mathematical induction $P\left( n \right)$ is true for all $n \in N$.
Hence proved.

Note:-Whenever we come up with this type of problem then there is also an alternate method to prove the result for that first, we had to solve the given equation and after that we had to use the expansion formula of ${\left( {a - b} \right)^x}$ by combination. After solving the expansion, we can take 25 common from the whole equation which will prove that the given equation is divisible by 25. But this will not be the easiest method because there are many chances of mistakes.