Question -
Answer -
Let P (n): 72n + 23n – 3. 3n – 1is divisible by 25
Let us check for n = 1,
P (1): 72 + 20.30
: 49 + 1
: 50
P (n) is true for n = 1. Where, P (n) is divisible by 25
Now, let us check for P (n) is true for n = k, and have toprove that P (k + 1) is true.
P (k): 72k + 23k – 3. 3k – 1 isdivisible by 25
: 72k + 23k – 3. 3k – 1 =25λ … (i)
We have to prove that:
72k + 1 + 23k. 3k isdivisible by 25
72k + 2 + 23k. 3k =25μ
So,
= 72(k + 1) + 23k. 3k
= 72k.71 + 23k. 3k
= (25λ – 23k – 3. 3k – 1) 49 + 23k.3k by using equation (i)
= 25λ. 49 – 23k/8. 3k/3. 49 + 23k.3k
= 24×25×49λ – 23k . 3k .49 + 24 . 23k.3k
= 24×25×49λ – 25 . 23k. 3k
= 25(24 . 49λ – 23k. 3k)
= 25μ
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.