If n is a positive integer, then n3 + 2n is divisible by(a) 2 (b) 6

1.	If n is a positive integer, then n3 + 2n is divisible by(a) 2 (b) 6 (c) 15 (d) 3
Answer» Answer: (D) = 3 For n = 1, n³ + 2n = 1 + 2 = 3 which is divisible by 3 and none of the other given alternatives. ∴ We shall prove n³+ 2n divisible by 3 for all n∈N. Let T(n) = n³+ 2n is divisible by 3. Basic Step: For n = 1, T(1) = n³ + 2n = 1 + 2 = 3 is divisible by 3 is true. Induction Step: Assume T(k) to be true, i.e., T(k) = k³+ 2k is divisible by 3 = k³ + 2k = 3m, where m∈N. ...(i) Now we need to prove that T(k + 1) holds true, i.e., (k + 1)³ + 2(k + 1) is divisible by 3. (k + 1)³ + 2(k + 1) = k³ + 3k² + 3k +1 + 2k + 2 = (k³ + 2k) + (3k² + 3k + 3) = 3m + 3 (k² + k + 1) (From (i)) ⇒ T(k + 1) = (k + 1)³ + 2 (k + 1) is divisible by 3, whenever T(k) = k³ + 2k is divisible by 3. ⇒ n³+ 2n is divisible by 3 ∀ n∈N.

If n is a positive integer, then n3 + 2n is divisible by(a) 2 (b) 6 (c) 15 (d) 3

Answer»

Answer: (D) = 3

For n = 1, n³ + 2n = 1 + 2 = 3 which is divisible by 3 and none of the other given alternatives.

∴ We shall prove n³+ 2n divisible by 3 for all n∈N.

Let T(n) = n³+ 2n is divisible by 3.

Basic Step:

For n = 1, T(1) = n³ + 2n = 1 + 2 = 3 is divisible by 3 is true.

Induction Step:

Assume T(k) to be true, i.e., T(k) = k³+ 2k is divisible by 3

= k³ + 2k = 3m, where m∈N. ...(i)

Now we need to prove that T(k + 1) holds true, i.e.,

(k + 1)³ + 2(k + 1) is divisible by 3.

(k + 1)³ + 2(k + 1) = k³ + 3k² + 3k +1 + 2k + 2

= (k³ + 2k) + (3k² + 3k + 3)

= 3m + 3 (k² + k + 1) (From (i))

⇒ T(k + 1) = (k + 1)³ + 2 (k + 1) is divisible by 3, whenever T(k) = k³ + 2k is divisible by 3.

⇒ n³+ 2n is divisible by 3 ∀ n∈N.