CBSE Class 11-Mathematics: Mathematical Induction Assignments (For CBSE, ICSE, IAS, NET, NRA 2022)

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Prove the Following using Principle of Mathematical induction

1) Prove that for any positive integer number is divisible by

2) Prove that

for all positive integers n.

3) For every is divisible by .

4) Prove by induction that

5) For every is multiple of

6) For every

7) For all

Where

8) If , then

9) Prove that for every .

10) Prove that

11) For all is divisible by .

Solution to Problem 1:

Let Statement is defined by

is divisible by

Step 1: Basic Step

We first show that p (1) is true. Let and calculate

is divisible by

Hence p (1) is true.

STEP 2: Inductive Hypothesis

We now assume that is true

is divisible by is equivalent to

, where B is a positive integer.

Step 3: Inductive Steps

We now consider the algebraic expression ; expand it and group like terms

Hence is also divisible by and therefore statement is true.

Solution to Problem 2:

Statement is defined by

Step 1: Basic Step

We first show that is true.

Left Side

Right Side

Hence is true.

STEP 2: Inductive Hypothesis

We now assume that is true

Step 3: Inductive Steps

Add to both sides

Factor on the right side

Set to common denominator and group

We have started from the statement and have shown that

Which is the statement .

Solution to Problem 3:

Let

is divisible by

Step 1: Basic Step

is just that is divisible by , which is trivial.

STEP 2: Inductive Hypothesis

We now assume that is true

i.e.. ,

is divisible by

Step 3: Inductive Steps

We must prove

Now

The first term is divisible by since is true and the second term is a multiple of . Hence, the last quantity is divisible by

Solution to Problem 4:

Statement is defined by

Step 1: Basic Step

We first show that is true.

Left Side

Right Side

Hence is true.

STEP 2: Inductive Hypothesis

We now assume that is true

Step 3: Inductive Steps

We must prove

Now

Taking LHS

Which is the statement

Solution to Problem 6:

Let Statement is defined by

For all

Step 1: Basic Step

Let

So is true

STEP 2: Inductive Hypothesis

We now assume that is true

That is,

Step 3: Inductive Steps

Let .

Then:

Now

So

> 3k + 1

Then holds for , and thus for all

Solution to Problem 7:

Let Statement is defined by

Where

Step 1: Basic Step

Let

Which is true

So is true

STEP 2: Inductive Hypothesis

We now assume that is true

Step 3: Inductive Steps

Let .

Then:

Taking the LHS

Now from hypothesis we know that

Also

So

Now kx2 is a positive quantity so we can say that

Which is

Solution to Problem 11:

Let Statement is defined by

For all is divisible by .

Step 1: Basic Step

Let .

Then the expression evaluates to , which is clearly divisible by 5.

STEP 2: Inductive Hypothesis

We now assume that is true

That is, that is divisible by .

Step 3: Inductive Steps

Let .

Then:

The first term in has as a factor (explicitly) , and the second term is divisible by (by assumption) . Since we can factor a out of both terms, then the entire expression, , must be divisible by .

Then holds for , and thus for all .