By the principle of mathematical induction, prove 4 + 8 + 12 + … + 4

1.	By the principle of mathematical induction, prove 4 + 8 + 12 + … + 4n = 2n(n + 1), for all n ∈ N.
Answer» Let P(n) denote the statement 4 + 8 + … + 4n = 2n(n + 1) i.e., P(n) : 4 + 8 + 12 + … + 4n = 2n(n + 1) Put n = 1, P(1): LHS = 4 RHS = 2 (1)(1 + 1) = 4 P(1) is true. Assume that P(n) is true for n = k P(k): 4 + 8 + 12 + … + 4k = 2k(k + 1) To prove P(k + 1) i.e., to prove 4 + 8 + 12 + … + 4k + 4(k + 1) = 2(k + 1) (k + 1 + 1) 4 + 8 + 12 + … + 4k + (4k + 4) = 2(k + 1) (k + 2) Consider, 4 + 8 + 12 + … + 4k + (4k + 4) = 2k(k + 1) + (4k + 4) = 2k(k + 1) + 4(k + 1) = 2k² + 2k + 4k + 4 = 2k² + 6k + 4 = 2(k + 1)(k + 2) P(k + 1) is also true. ∴ By Mathematical Induction, P(n) for all value n ∈ N.

By the principle of mathematical induction, prove 4 + 8 + 12 + … + 4n = 2n(n + 1), for all n ∈ N.

Answer»

Let P(n) denote the statement 4 + 8 + … + 4n = 2n(n + 1)

i.e., P(n) : 4 + 8 + 12 + … + 4n = 2n(n + 1)

Put n = 1,

P(1): LHS = 4

RHS = 2 (1)(1 + 1) = 4

P(1) is true.

Assume that P(n) is true for n = k

P(k): 4 + 8 + 12 + … + 4k = 2k(k + 1)

To prove P(k + 1)

i.e., to prove 4 + 8 + 12 + … + 4k + 4(k + 1) = 2(k + 1) (k + 1 + 1)

4 + 8 + 12 + … + 4k + (4k + 4) = 2(k + 1) (k + 2)

Consider, 4 + 8 + 12 + … + 4k + (4k + 4) = 2k(k + 1) + (4k + 4)

= 2k(k + 1) + 4(k + 1)

= 2k² + 2k + 4k + 4

= 2k² + 6k + 4

= 2(k + 1)(k + 2)

P(k + 1) is also true.

∴ By Mathematical Induction, P(n) for all value n ∈ N.