2026/086) Prove that for every positive integer n, the number $A_n=7^{2n}−48n−1$ is a multiple of 9.

Let us take the mod 9 of the expression

$A_n=7^{2n} - 48n -1 \mod 9$

 $=(7^2)^n - 48n -1 \mod 9$

$=49^n - 48n -1 \mod 9$ 

$=4^n -3n -1 \mod 9$ reducing to mod 9

This can be proved by principle of mathematical induction  also binomial expansion

Method 1 Principle of Mathematical Induction 

To prove by   principle of mathematical induction we first need to check for initial condition then induction step  

We have 

$A_1= 4^1 - 3 *1 -1 = 0$  is divisible by 9

We have checked the basic step

So we proceed to check the induction step

 Let us assume that it is true for some n say k 

So  $A_k= 4^k - 3k -1 \equiv 0 \mod 9$

We need to prove

  $A_{k+1}= 4^{k+1} - 3 *(k+1) -1 \equiv 0 \mod 9\cdots(2)$

We have

$A_k= 4^k - 3k -1$

$A_{k+1}= 4^{k+1} - 3(k+1) -1$ 

Now $A_{k+1}- 4 * A_k = 4^{k+1} - 3(k+1) -1- 4 (4^k - 3k -1)$

$= -3k -4 + 12k + 4 = 9k$

Hence Or $A_{k+1} - 4 * A_k+9k$

or  $A_{k+1} = 4 * A_k+9k$

 So if $A_k$ is divisible by 9 the $A_{k+1}$ is divisible by 9

We proved the induction step

Hence Proved

Method 2 :Binomial expansion

We shall prove the same by Binomial expansion

Let us use for 1 and 2 for simple cases and the binomial theorem for $n>2$

$A_k = 4^k - 3k -1$

So we have

$A_1 = 4 - 3 - 1 =0$ or $A_1$ is divisible by 9

and $A_2 = 4^2 - 3 * 2 -1 = 9$ or $A_2$ is divisible by 9

 for $n>2$ we have (it is rue for all n but we shall use for n>2)

$4^n= (3+1)^n =  \sum_{k=0}^n \binom {n}{k} 3^{n-k}1^k$

 $=\sum_{k=0}^{n-2} \binom {n}{k} 3^{n-k}1^k+3n + 1$ by grouping 1st n-2 terms and separating last 2 terms

As the sum under the summation each term is divisible by n we get

$4^n = 9p + 3n +1 $ for sum mutilple of 9 with some p

Or $4^n -3n -1 = 9p$

Or   $4^n -3n -1$ s divisible by 9

Hence proved