For all natural numbers n, the expression 2.7n + 3.5n – 5 is divisib

1.	For all natural numbers n, the expression 2.7n + 3.5n – 5 is divisible by(a) 16 (b) 24 (c) 20 (d) 21
Answer» Answer : (b) = 24 When n = 1, 2.7ⁿ + 3.5ⁿ – 5 = 2.7 + 3.5 – 5 = 24 which is divisible by 24 and none of the other given alternatives. ∴ We need to prove 2.7ⁿ+ 3.5ⁿ – 5 is divisible by 24 ∀ n∈N. Let T(n) be the statement 2.7ⁿ+ 3.5ⁿ– 5 is divisible by 24. T(1) holds true as shown above. Assume T(k) to be true, i.e., 2.7^k + 3.5^k – 5 is divisible by 24, i.e., 2.7^k + 3.5^k – 5 = 24m, m∈N ....(i) Now 2.7^{k + 1} + 3.5^{k + 1}– 5 = 2.7.7^k + 3.5.5^k – 5 = (2.7^k + 3.5^k– 5) + 12(7^k ) + 12 (5^k ) = 24m + 12 (7^k + 5^k ) \(\big[Now\,7^k\,and\,5^k\,,k\,\in\,N\,being \,both\,odd\,,their\,sum\,is\,even. Let\,7^k\,+5^k\,=2x\,,\,x\,\in\,N\big]\) = 24m + 12 (2x) ; m, x∈N = 24 (m + x) ⇒ 2.7^{k + 1} + 3.5^{k + 1} – 5 is divisible by 24 ⇒ T (k + 1) is true whenever T(k) is true, k∈N. ⇒ 2.7ⁿ + 3.5ⁿ – 5 is divisible by 24 for all n∈N.

For all natural numbers n, the expression 2.7n + 3.5n – 5 is divisible by(a) 16 (b) 24 (c) 20 (d) 21

Answer»

Answer : (b) = 24

When n = 1,

2.7ⁿ + 3.5ⁿ – 5 = 2.7 + 3.5 – 5 = 24 which is divisible by 24 and none of the other given alternatives.

∴ We need to prove 2.7ⁿ+ 3.5ⁿ – 5 is divisible by 24 ∀ n∈N.

Let T(n) be the statement 2.7ⁿ+ 3.5ⁿ– 5 is divisible by 24.

T(1) holds true as shown above.

Assume T(k) to be true, i.e., 2.7^k + 3.5^k – 5 is divisible by 24, i.e.,

2.7^k + 3.5^k – 5 = 24m, m∈N ....(i)

Now 2.7^{k + 1} + 3.5^{k + 1}– 5 = 2.7.7^k + 3.5.5^k – 5

= (2.7^k + 3.5^k– 5) + 12(7^k ) + 12 (5^k )

= 24m + 12 (7^k + 5^k )

\(\big[Now\,7^k\,and\,5^k\,,k\,\in\,N\,being \,both\,odd\,,their\,sum\,is\,even. Let\,7^k\,+5^k\,=2x\,,\,x\,\in\,N\big]\)

= 24m + 12 (2x) ; m, x∈N = 24 (m + x)

⇒ 2.7^{k + 1} + 3.5^{k + 1} – 5 is divisible by 24

⇒ T (k + 1) is true whenever T(k) is true, k∈N.

⇒ 2.7ⁿ + 3.5ⁿ – 5 is divisible by 24 for all n∈N.