CBSE Class 11 Mathematics Chapter 4: Principle of Mathematical Induction NCERT Solutions

Let the given statement be denoted by P(n). So, P(n) is:

1+3+3^2+...+3^{n-1}=\frac{\left(3^n-1\right)}{2}

Step 1: Verify the base case.

For n = 1, the left-hand side (LHS) is the first term, which is 1.

The right-hand side (RHS) is \frac{\left(3^1-1\right)}{2} = \frac{3-1}{2} = \frac{2}{2} = 1.

Since LHS = RHS, P(1) is true.

Step 2: Assume the inductive hypothesis.

Assume that P(k) is true for some positive integer k. That is,

1+3+3^2+...+3^{k-1} = \frac{\left(3^k-1\right)}{2} ... (i)

Step 3: Prove the inductive step.

We need to prove that P(k+1) is true. P(k+1) is the statement:

1+3+3^2+...+3^{(k+1)-1} = \frac{\left(3^{k+1}-1\right)}{2}

Consider the LHS of P(k+1):

1+3+3^2+...+3^{k-1}+3^{(k+1)-1}

This can be written as:

(1+3+3^2+...+3^{k-1}) + 3^k

Using the inductive hypothesis (i), we can substitute the sum up to 3^{k-1}:

\frac{\left(3^k-1\right)}{2} + 3^k

Now, let's simplify this expression:

= \frac{\left(3^k-1\right) + 2 \cdot 3^k}{2}

= \frac{3^k - 1 + 2 \cdot 3^k}{2}

= \frac{(1+2) \cdot 3^k - 1}{2}

= \frac{3 \cdot 3^k - 1}{2}

= \frac{3^{k+1}-1}{2}

This is the RHS of P(k+1). Thus, P(k+1) is true whenever P(k) is true.

Conclusion:

By the Principle of Mathematical Induction, the statement P(n) is true for all natural numbers n \in \mathbb{N}.

Let the given statement be denoted by P(n). So, P(n) is:

1^3 + 2^3 + 3^3 + ... + n^3 = \left(\frac{n(n+1)}{2}\right)^2

Step 1: Verify the base case.

For n = 1, the LHS is 1^3 = 1.

The RHS is \left(\frac{1(1+1)}{2}\right)^2 = \left(\frac{1 \cdot 2}{2}\right)^2 = \left(\frac{2}{2}\right)^2 = 1^2 = 1.

Since LHS = RHS, P(1) is true.

Step 2: Assume the inductive hypothesis.

Assume that P(k) is true for some positive integer k. That is,

1^3 + 2^3 + 3^3 + ... + k^3 = \left(\frac{k(k+1)}{2}\right)^2 ... (i)

Step 3: Prove the inductive step.

We need to prove that P(k+1) is true. P(k+1) is the statement:

1^3 + 2^3 + 3^3 + ... + k^3 + (k+1)^3 = \left(\frac{(k+1)((k+1)+1)}{2}\right)^2 = \left(\frac{(k+1)(k+2)}{2}\right)^2

Consider the LHS of P(k+1):

(1^3 + 2^3 + 3^3 + ... + k^3) + (k+1)^3

Using the inductive hypothesis (i), we can substitute the sum of the first k cubes:

\left(\frac{k(k+1)}{2}\right)^2 + (k+1)^3

Now, let's simplify this expression:

= \frac{k^2(k+1)^2}{4} + (k+1)^3

Factor out the common term (k+1)^2:

= (k+1)^2 \left[ \frac{k^2}{4} + (k+1) \right]

Find a common denominator inside the bracket:

= (k+1)^2 \left[ \frac{k^2 + 4(k+1)}{4} \right]

= (k+1)^2 \left[ \frac{k^2 + 4k + 4}{4} \right]

Recognize the perfect square in the numerator:

= (k+1)^2 \left[ \frac{(k+2)^2}{4} \right]

Combine the terms:

= \frac{(k+1)^2 (k+2)^2}{4}

= \left(\frac{(k+1)(k+2)}{2}\right)^2

This is the RHS of P(k+1). Thus, P(k+1) is true whenever P(k) is true.

Conclusion:

By the Principle of Mathematical Induction, the statement P(n) is true for all natural numbers n \in \mathbb{N}.