Proof by Mathematical Induction


Using the principle to proof by mathematical induction we need to follow the techniques and steps exactly as shown.

We note that a prove by mathematical induction consists of three steps.

• Step 1. (Basis) Show that P(n₀) is true.

• Step 2. (Inductive hypothesis). Write the inductive hypothesis: Let k be an integer such that k ≥ n₀ and P(k) be true.

• Step 3. (Inductive step). Show that P(k + 1) is true.


Questions with solutions to Proof by Mathematical Induction

1. Using the principle of mathematical induction, prove that
a + ar + ar2 + ....... + arn – 1 = (arn – 1)/(r - 1) for r > 1 and all n ∈ N.

Solution:

Let the given statement be P(n). Then,

P(n): a + ar + ar2 + ….... +arn - 1 = {a(rn -1)}/(r - 1).

When n = 1, LHS = a and RHS = {a(r1 - 1)}/(r - 1) = a

Therefore LHS = RHS.

Thus, P(1) is true.

Let P(k) be true. Then,

P(k): a + ar + ar2 + …… + ark - 1 = {a(rk - 1)}/(r - 1)

Now, (a + ar + ar2 + …... + ark - 1) + ark = {a(rk - 1)}/(r - 1) + ar2                                                                      ....... [using(i)]
                             = a(rk + 1 - 1)/(r - 1).

Therefore,
P(k + 1): a + ar + ar2 + …….. +ark - 1 + ark = {a(rk + 1 - 1)}/(r - 1)

⇒ P(k + 1)is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

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





Proof by Mathematical Induction

2. Let a and b be arbitrary real numbers. Using the principle of mathematical induction, prove that
(ab)n = anbn for all n ∈ N.

Solution:

Let the given statement be P(n). Then,

P(n): (ab)n = anbn.

When = 1, LHS = (ab)1 = ab and RHS = a1b1 = ab

Therefore LHS = RHS.

Thus, the given statement is true for n = 1, i.e., P(1) is true.

Let P(k) be true. Then,

P(k): (ab)k = akbk.

Now, (ab)k + 1 = (ab)k (ab)

                             = (akbk)(ab) [using (i)]

                              = (ak ∙ a)(bk ∙ b) [by commutativity and associativity of multiplication on real numbers]

                              = (ak + 1 ∙ bk + 1 ).

Therefore P(k+1): (ab)k + 1 = ((ak + 1 ∙ bk + 1)

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

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



More examples to Proof by Mathematical Induction

3. Using the principle of mathematical induction, prove that (xn - yn) is divisible by (x - y)for all n ∈ N.

Solution:

Let the given statement be P(n). Then,

P(n): (xn - yn) is divisible by (x - y).

When n = 1, the given statement becomes: (x1 - y1) is divisible by (x - y), which is clearly true.

Therefore P(1) is true.

Let p(k) be true. Then,

P(k): xk - yk is divisible by (x-y).

Now, xk + 1 - yk + 1 = xk + 1 - xky - yk + 1

                              [on adding and subtracting x)ky]

= xk(x - y) + y(xk - yk), which is divisible by (x - y) [using (i)]

⇒ P(k + 1): xk + 1 - yk + 1is divisible by (x - y)

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the Principal of Mathematical Induction, P(n) is true for all n ∈ N.



4. Using the principle of mathematical induction, prove that (102n - 1 + 1) is divisible by 11 for all n ∈ N.

Solution:

Let P (n): (102n – 1 + 1) is divisible by 11.

For n=1, the given expression becomes {10(2 × 1 - 1) + 1} = 11, which is divisible by 11.

So, the given statement is true for n = 1, i.e., P (1) is true.

Let P(k) be true. Then,

P(k): (102k - 1 + 1) is divisible by 11

⇒ (102k - 1 + 1) = 11 m for some natural number m.

Now, {102(k - 1) - 1 + 1} = (102k + 1 + 1) = {102 ∙ 10(2k - 1)+ 1}

                              = 100 × {102k - 1+ 1 } - 99

                              = (100 × 11m) - 99

                              = 11 × (100m - 9), which is divisible by 11

⇒ P (k + 1): {102(k + 1) - 1 + 1} is divisible by 11

⇒ P (k + 1) is true, whenever P(k) is true.

Thus, P (1) is true and P(k + 1) is true , whenever P(k) is true.

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



5. Using the principle if mathematical induction, prove that (7n – 3n) is divisible by 4 for all n ∈ N.

Solution:

Let P(n) : (7n – 3n) is divisible by 4.

For n = 1, the given expression becomes (7 1 - 3 1) = 4, which is divisible by 4.

So, the given statement is true for n = 1, i.e., P(1) is true.

Let P(k) be true. Then,

P(k): (7k - 3k) is divisible by 4.

⇒ (7k - 3k) = 4m for some natural number m.

Now, {7(k + 1) - 3 (k + 1)} = 7(k + 1) – 7 ∙ 3k + 7 ∙ 3k - 3 (k + 1)
                                           (on subtracting and adding 7 ∙ 3k)

                              = 7(7k - 3k) + 3 k (7 - 3)

                              = (7 × 4m) + 4 ∙ 3k

                              = 4(7m + 3k), which is clearly divisible by 4.

∴ P(k + 1): {7(k + 1) - 3 (k + 1)} is divisible by 4.

⇒ P(k + 1) is true, whenever P(k) is true.

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



Solved examples to Proof by Mathematical Induction

6. Using the principle if mathematical induction, prove that
(2 ∙ 7n + 3 ∙ 5n - 5) is divisible by 24 for all n ∈ N.

Solution:

Let P(n): (2 ∙ 7n + 3 ∙ 5n - 5) is divisible by 24.

For n = 1, the given expression becomes (2 ∙ 71 + 3 ∙ 51 - 5) = 24, which is clearly divisible by 24.

So, the given statement is true for n = 1, i.e., P(1) is true.

Let P(k) be true. Then,

P(k): (2 ∙ 7n + 3 ∙ 5n - 5) is divisible by 24.

⇒ (2 ∙ 7n + 3 ∙ 5n - 5) = 24m, for m = N

Now, (2 ∙ 7n + 3 ∙ 5n - 5)

                     = (2 ∙ 7k ∙ 7 + 3 ∙ 5k ∙ 5 - 5)

                     = 7(2 ∙ 7k + 3 ∙ 5k - 5) = 6 ∙ 5k + 30

                     = (7 × 24m) - 6(5k - 5)

                     = (24 × 7m) - 6 × 4p, where (5k - 5) = 5(5k - 1 - 1) = 4p
                                        [Since (5k - 1 - 1) is divisible by (5 - 1)]

                     = 24 × (7m - p)

                     = 24r, where r = (7m - p) ∈ N

⇒ P (k + 1): (2 ∙ 7k + 13 ∙ 5k + 1 - 5) is divisible by 24.

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

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


 Mathematical Induction

Mathematical Induction

Problems on Principle of Mathematical Induction

Proof by Mathematical Induction

Induction Proof




11 and 12 Grade Math 

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