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Mathematics

Principle of Mathematical Induction

Gk Scientist April 19, 2022 No Comments
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Principle of Mathematical Induction:

Often it becomes a difficult job to prove a mathematical statement by a direct method. In such a case, an indirect method, known as the method of mathematical induction is used. This is a special technique to prove a mathematical statement involving a variable n, usually denoting a positive integer, in three steps. The steps are-

Verification Step- At this stage, the mathematical statement is verified to be true for the smallest permissible value of n.

Assumption Step- Here, the statement is assumed to be true for n = m, where m is any positive integer greater than the permissible lowest value of n.

Induction Step- This may also be referred to as the concluding step where the statement, which has been assumed to be true for n = m, is shown to be true for n = m + 1 only.

Now, as the statement has already been verified for n = 1 (or for the lowest permissible value of n), following the induction step, it will be true for n = 2 and so on. Thus, logically the statement is proved to be true for any value of n.

Hence, the principle of mathematical induction states that “If P(n) be a statement such that P(1) is verified to be true and P(m + 1) is true whenever P(m) is true, m being a positive integer, then the statement is true for all positive integral values of n.

Example- Use the principle of mathematical induction to prove the validity of the following statements for all n ∈ N.

(a) n2 + n is an even number.

Solution- Let P(n) be the statement “n2 + n is an even number”.

Put n = 1
P(1) = (1)2 + 1 = 2 = an even number
∴ P(n) is true for n = 1.

Suppose P(n) is true for n = k i.e. k2 + k is an even number.
Let k2 + k = 2λ
⇒ k2 = 2λ – k ………..(i)

Now, we have to show that P(k+1) is true whenever P(k) is true i.e. we have to show that (k + 1)2 + (k + 1) is an even number.

∴ (k + 1)2 + (k + 1)
= k2 +2k + 1 + k + 1
= k2 + 3k + 2
= 2λ – k + 3k + 2 [using (i)]
= 2λ + 2k + 2
= 2 [λ + k + 1]
= 2μ, μ ∈ N
= an even number

Therefore, P (k + 1) is true whenever P(k) is true.
Hence by the principle of mathematical induction, P(n) is true if n ∈ N.

(b) 10n + 3×4n+2 + 5 is divisible by 9.

Solution- Let P(n) be the statement “10n + 3×4n+2 + 5 is divisible by 9″.

Put n = 1
P(1) = 101 + 3×41+2 + 5 = 10 + 192 + 5 = 207 = 9 x 23 = divisible by 9
∴ P(n) is true for n = 1.

Suppose P(n) is true for n = k i.e. 10k + 3×4k+2 + 5 is divisible by 9.
Let 10k + 3×4k+2 + 5 = 9λ
⇒ 10k = 9λ – 3×4k+2 – 5 …………(i)

Now, we have to show that P(k+1) is true whenever P(k) is true i.e. we have to show that 10k+1 + 3×4(k+1) + 2 + 5 is divisible by 9.

∴ 10k+1 + 3×4(k+1) + 2 + 5
= 10k . 10 + 3×4(k+2) . 41 + 5
= (9λ – 3×4k+2 – 5) 10 + 12×4k+2 + 5
= 90λ – 30×4k+2 -50 + 12×4k+2 + 5
= 90λ – 18×4k+2 -45
= 9 (10λ – 2×4k+2 – 5)
= 9μ is divisible by 9

Therefore, P (k + 1) is true whenever P(k) is true.
Hence by the principle of mathematical induction, P(n) is true if n ∈ N.

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