0 votes
73 views
in Mathematical Induction by (52.3k points)
closed by

Using the principle of mathematical induction, prove each of the following for all n ϵ N: 

(32n+2 – 8n – 9) is divisible by 8.

1 Answer

+1 vote
by (56.6k points)
selected by
 
Best answer

To Prove:

(32n+2 – 8n – 9) is divisible by 8.

Let us prove this question by principle of mathematical induction (PMI) for all natural numbers

Let P(n): (32n+2 – 8n – 9) is divisible by 8

For n = 1 P(n) is true since

32n+2 – 8n – 9 = 32+2 - 8 x 1 - 9 = 81 - 17 = 64

Which is divisible of 8 

Assume P(k) is true for some positive integer k , ie,

= (32k+2 – 8k – 9) is divisible by 8

∴ 32k+2 – 8k – 9 = m x 8, where m ∈ N …(1)

We will now prove that P(k + 1) is true whenever P( k ) is true 

Consider,

[Adding and subtracting 8k + 9]

= 8×r, where r = 9m + 8k + 8 is a natural number 

Therefore 32k+2 - 8k - 9 is a divisible of 8 

Therefore, P (k + 1) is true whenever P(k) is true 

By the principle of mathematical induction, P(n) is true for all natural numbers, ie, N. 

Hence proved.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...