Question

use principle of mathematical induction to prove the above question​

Answer 1

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]

Now, divide the whole equation by 25

we will get,

(-1)ⁿ + (-1)ⁿ⁻¹ = 0

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Answer 2

Checking whether P(1) is true.

P(1) = 7² + 2⁰.3⁰ = 49 + 1 = 50

50 is divisible by 25.

So P(1) is true.

Let P(K) be true. Then,

P(K) = $\sf{7^{2K}+3^{3k-3}.3^{k-1}=25x}$

Where x is an integer.

Now check if P(K+1) is True.

P(K + 1) = $\sf{7^{2(K+1)}+2^{3(K+1)-3}\times 3^{(K+1)-1}}$

$\sf{=(7^2^K\times49)+2^{3K-3}\times8\times3^{k-1}\times3}$

$\sf{=49\times7^2^K+24[2^{3K-3}3^{K+1}]}$

$\sf{= 49\times7^{2K}+24[25m-7^{2K}]}$

$\sf{= 49\times7^{2K}-24\times7^{2K}+24\times25x}$

$\sf{=25[7^{2K}+24m]}$

$\sf{=25\times x}$

Where x is an integer.

So, the given expression is divisible by 25 and is thus, true.

∴ P(K + 1) is true whenever P(K) is true. So using Principal of Mathematical Induction, we can conclude P(n) is true for all  n∈N.