Question

Use the principle of mathematical induction to prove that
4+ 8 + 12 + ... + 4 n=2 n(n+1) VnEN.​

Answer 1

$\underline{\underline{\blue{\textbf{Step - By - Step - Explanation : -}}}}$

To prove :

• 4 + 8 + 12 + ...+ 4n = 2n(n + 1)

proof :

Let the Given statement be p(n)

p(n): 4 + 8 + 12 + ...+4n = 2n(n + 1)

For n = 1

P(1) : 4 = 2 × 1(1 + 1) = 2 × 2 = 4

So,

p(1) is true when n = 1 .

Now, Let us assume that p(n) is true for some positive intiger k

p(k): 4 + 8 + 12 +...+ 4k = 2k(k + 1) ...(1)

Now , we need to prove that p(k + 1) is also true.

p(k + 1): 4 + 8 + 12 + .... + 4k + 4(k +1) = 2(k + 1)(k + 2)

LHS :

= 4 + 8 + 12 + ...+ 4k + 4(k + 1)

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

= 2k² + 2k + 4k + 4

= 2k² + 6k + 4

RHS :

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

= 2[k(k + 2) + 1(k + 2)]

= 2[k² + 2k + k + 2]

= 2(k² + 3k + 2)

= 2k² + 6k + 4

So,

p(k + 1) is true when p(k) is true.

$\large\dag \large { \red{\underline{\bf{Hence }}}}$By principal of mathematical induction, statement p(n) is true for all positive intigers.

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