Using binomial theorem prove that 6^n-5n always leaves reminder 1 when divided by 25?

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Proof that 6^n - 5n always leaves a remainder of 1 when divided by 25 using binomial theorem:

To prove this statement, we can use the binomial theorem which states that for any positive integer n:

(a + b)^n = C(n, 0)a^n b^0 + C(n, 1)a^(n-1) b^1 + C(n, 2)a^(n-2) b^2 + ... + C(n, n-1)a^1 b^(n-1) + C(n, n)a^0 b^n

where C(n, k) represents the binomial coefficient of n and k.

Now, let's consider the expression 6^n - 5n. We want to prove that it always leaves a remainder of 1 when divided by 25.

Step 1: Expand (6 + (-5))^n using the binomial theorem.

The binomial theorem tells us that:

(6 + (-5))^n = C(n, 0)(6^n) (-5)^0 + C(n, 1)(6^(n-1)) (-5)^1 + C(n, 2)(6^(n-2)) (-5)^2 + ... + C(n, n-1)(6^1) (-5)^(n-1) + C(n, n)(6^0) (-5)^n

Simplifying this expression, we get:

6^n - 5n = C(n, 0)(6^n) + C(n, 1)(6^(n-1)) (-5) + C(n, 2)(6^(n-2)) (-5)^2 + ... + C(n, n-1)(6^1) (-5)^(n-1) + C(n, n)(-5)^n

Step 2: Rearrange the terms.

We can rearrange the terms in the expression as follows:

6^n - 5n = [C(n, 0)(6^n) + C(n, 1)(6^(n-1)) (-5) + C(n, 2)(6^(n-2)) (-5)^2 + ... + C(n, n-1)(6^1) (-5)^(n-1)] + C(n, n)(-5)^n

Step 3: Observe the terms inside the square brackets.

Notice that the terms inside the square brackets are all multiples of 25. This is because each term in the expansion contains a factor of (-5)^k, where k is a positive integer.

Step 4: Simplify the expression.

Since the terms inside the square brackets are all multiples of 25, they can be expressed as 25k, where k is an integer. Therefore, we have:

6^n - 5n = 25k + C(n, n)(-5)^n

Step 5: Prove that the remainder is always 1.

We need to show that C(n, n)(-5)^n leaves a remainder of 1 when divided by 25.

Now, consider the binomial coefficient C(n, n). It represents the number

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