Prove by the principle of mathematical induction:72n + 23n – 3. 3n

1.	Prove by the principle of mathematical induction:72n + 23n – 3. 3n – 1 is divisible by 25 for all n ϵ N
Answer» Suppose P (n): 7²ⁿ + 2^{3n – 3}. 3n – 1 is divisible by 25 Now 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 Then, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true. P (k): 7^2k + 2^{3k – 3}. 3k – 1 is divisible by 25 : 7^2k + 2^{3k – 3}. 3^{k – 1} = 25λ … (i) Now we have to prove that: 7^{2k + 1} + 2^3k. 3^k is divisible by 25 7^{2k + 2} + 2^3k. 3^k = 25μ Therefore, = 7^{2(k + 1)} + 2^3k. 3^k = 7^2k.7¹ + 2^3k. 3^k = (25λ – 2^{3k – 3}. 3^{k – 1}) 49 + 2^3k. 3k by using equation (i) = 25λ. 49 – 2^3k/8. 3^k/3. 49 + 2^3k. 3^k = 24 × 25 × 49λ – 2^3k. 3^k. 49 + 24 . 2^3k.3^k = 24 × 25 × 49λ – 25 . 2^3k. 3^k = 25(24 . 49λ – 2^3k. 3^k) = 25μ P (n) is true for n = k + 1 Thus, P (n) is true for all n ∈ N.

Prove by the principle of mathematical induction:72n + 23n – 3. 3n – 1 is divisible by 25 for all n ϵ N

Answer»

Suppose P (n): 7²ⁿ + 2^{3n – 3}. 3n – 1 is divisible by 25

Now 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

Then, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.

P (k): 7^2k + 2^{3k – 3}. 3k – 1 is divisible by 25

: 7^2k + 2^{3k – 3}. 3^{k – 1} = 25λ … (i)

Now we have to prove that:

7^{2k + 1} + 2^3k. 3^k is divisible by 25

7^{2k + 2} + 2^3k. 3^k = 25μ

Therefore,

= 7^{2(k + 1)} + 2^3k. 3^k

= 7^2k.7¹ + 2^3k. 3^k

= (25λ – 2^{3k – 3}. 3^{k – 1}) 49 + 2^3k. 3k by using equation (i)

= 25λ. 49 – 2^3k/8. 3^k/3. 49 + 2^3k. 3^k

= 24 × 25 × 49λ – 2^3k. 3^k. 49 + 24 . 2^3k.3^k

= 24 × 25 × 49λ – 25 . 2^3k. 3^k

= 25(24 . 49λ – 2^3k. 3^k)

= 25μ

P (n) is true for n = k + 1

Thus, P (n) is true for all n ∈ N.