# Prove by method of induction, for all n ∈ N: [1201]n=[12n01] ∀ n ∈ N - Mathematics and Statistics

Sum

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

[(1, 2),(0, 1)]^"n" = [(1, 2"n"),(0, 1)] ∀ n ∈ N

#### Solution

Let P(n) ≡ [(1, 2),(0, 1)]^"n" = [(1, 2"n"),(0, 1)]

Step 1:

For n = 1,

L.H.S. = [(1, 2),(0, 1)]^1 = [(1, 2),(0, 1)]

R.H.S. = [(1, 2),(0, 1)]

∴ L.H.S. = R.H.S.

∴ P(1) is true.

Step 2:

Let us assume that for some k ∈ N, P(k) is true,

i.e., [(1, 2),(0, 1)]^"k" = [(1, 2"k"),(0, 1)]     ...(1)

Step 3:

To prove that P(k+ 1) is true, i.e., to prove that

[(1, 2),(0, 1)]^("k" + 1) = [(1, 2("k" + 1)),(0, 1)]

Now, L.H.S. = [(1, 2),(0, 1)]^("k" + 1)

= [(1, 2),(0, 1)]^"k" [(1, 2),(0, 1)]

= [(1, 2"k"),(0, 1)] [(1, 2),(0, 1)]    ...[By (1)]

= [(1 + 0, 2 + 2"k"),(0 + 0, 0 + 1)]

= [(1, 2("k" + 1)),(0, 1)]

= R.H.S.

∴ P(k + 1) is true.

Step 4:

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

i.e., [(1, 2),(0, 1)]^"n" = [(1, 2"n"),(0, 1)], ∀ n ∈ N

Concept: Principle of Mathematical Induction
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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 16 | Page 74