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# Prove by method of induction, for all n ∈ N: (24n−1) is divisible by 15 - Mathematics and Statistics

Sum

Prove by method of induction, for all n ∈ N:

(24n−1) is divisible by 15

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#### Solution

(24n−1) is divisible by 15 if and only if (24n−1) is a multiple of 15.

Let P(n) ≡ (24n−1) = 15m, where m ∈ N

Step I:

Put n = 1

∴ 24(1) − 1 = 16 − 1 = 15

∴ (24n − 1) is a multiple of 15.

∴ P(1) is true

Step II:

Let us consider that P(n) is true for n = k

i.e., 24k − 1 is a multiple of 15.

∴ 24k − 1 = 15a, where a ∈ N

∴ 24k = 15a + 1    ...(i)

Step III:

We have to prove that P(k + 1) is true

i.e., to prove that 2^(4("k"+1)) − 1 is a multiple of 15.

i.e., 2^(4("k"+1)) − 1 = 15b, where b ∈ N

∴ P(k + 1) = 2^(4("k"+1)) − 1 = 2^(4"k"+4) − 1

= 24k.24 − 1

= 16.(24k) − 1

= 16(15a + 1) − 1   ...[From (i)]

= 240a + 16 − 1

= 240a + 15

= 15(16a + 1)

= 15b, where b = (16a + 1) ∈ N

∴ P(k + 1) is true

Step IV:

From all steps above by the principle of mathematical induction, P(n) is true for all n ∈ N.

∴  (24n − 1) is divisible by 15, for all n ∈ N.

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#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 4 Methods of Induction and Binomial Theorem
Exercise 4.1 | Q 11 | Page 74
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