Question

Prove that 7^2n + 2^(3n-3).3^(n-1) is divisible by 25 for every n natural number

Answer 1

Let S(n) be the statement 72n+23(n−1)×3n−1=72n+23n−3×3n−1 is a multiple of 25, n∈N.

Basis step: S(1): 72(1)+23(1)−3×31−1=49+1×1=50, which is a multiple of 25.

Inductive step:

Assume S(k) is true, i.e. assume that 72k+23k−3×3k−1 is a multiple of 25, k∈N.

⇒72k+23k−3×3k−1=25A, A∈N

Then, S(k+1): 72(k+1)+23(k+1)−3×3(k+1)−1

=72k+2+23k×3k

=49×72k+23k×3k

=49×72k+23k−3×3k−1−23k−3×3k−1+23k×3k

=49×25A−23k−3×3k−1+23k×3k

=1225A−49×23k−3×3k−1+23k×3k

=1225A−49×23k−3×3k−1+8×23k−3×3×3k−1

=1225A−49×23k−3×3k−1+24×23k−3×3k−1

=1225A−25×23k−3×3k−1

=25(49A−23k−3×3k−1)

=25(49A−23(k−1)×3k−1), which is a multiple of 25.

So S(k+1) is true whenever S(k) is true.

Therefore, 72n+23(n−1)×3n−1 is a multiple of 25, n∈N.

Answer 2

7²ⁿ + 2³ⁿ.3ⁿ⁻¹

= (7²)ⁿ + (2³)ⁿ⁻¹. 3ⁿ⁻¹

= 49ⁿ  + 8ⁿ⁻¹.3ⁿ⁻¹

= (49)ⁿ + (8.3)ⁿ⁻¹

= 49ⁿ + 24ⁿ⁻¹

= (50-1)ⁿ + (25-1)ⁿ⁻¹

= 50ⁿ -n.50ⁿ⁻¹ + ....+(-1)ⁿ⁻¹. n50 +(-1)ⁿ] + [25ⁿ⁻¹ - n.25ⁿ⁻²+.... + (-1)ⁿ⁻². (n-1).25 +(-1)ⁿ⁻¹]

divide the whole equation by 25

we get

(-1)ⁿ + (-1)ⁿ⁻¹ = 0