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

Solution: Let us consider the statement P(n): \(3n < n!\).

We need to check the truth of this statement for n = 1, 2, 3, and then for n \ge 4.

For n = 1:

The statement becomes \(3 \times 1 < 1!\), which simplifies to \(3 < 1\). This is false.

For n = 2:

The statement becomes \(3 \times 2 < 2!\), which simplifies to \(6 < 2\). This is false.

For n = 3:

The statement becomes \(3 \times 3 < 3!\), which simplifies to \(9 < 6\). This is false.

For n = 4:

The statement becomes \(3 \times 4 < 4!\), which simplifies to \(12 < 24\). This is true.

For n = 5:

The statement becomes \(3 \times 5 < 5!\), which simplifies to \(15 < 120\). This is true.

Since the factorial function \(n!\) grows much faster than the linear function \(3n\), the inequality \(3n < n!\) will hold true for all integers \(n \ge 4\). Therefore, P(n): \(3n < n!\) is a statement that is true for all \(n \ge 4\) but not for n=1, 2, or 3.

Solution: Consider the statement P(n) which represents the sum of the squares of the first n natural numbers:

$$P(n):\; 1^2\; +\; 2^2\; +\; 3^2\; +\; ...\; +\; n^2\; =\; \backslash frac\{n(n+1)(2n+1)\}\{6\}$$

We can verify this statement for the first few natural numbers:

For n = 1:

Left Hand Side (LHS) = \(1^2 = 1\)

Right Hand Side (RHS) = \(\frac{1(1+1)(2\times 1+1)}{6} = \frac{1(2)(3)}{6} = \frac{6}{6} = 1\)

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

For n = 2:

LHS = \(1^2 + 2^2 = 1 + 4 = 5\)

RHS = \(\frac{2(2+1)(2\times 2+1)}{6} = \frac{2(3)(5)}{6} = \frac{30}{6} = 5\)

Since LHS = RHS, P(2) is true.

For n = 3:

LHS = \(1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14\)

RHS = \(\frac{3(3+1)(2\times 3+1)}{6} = \frac{3(4)(7)}{6} = \frac{84}{6} = 14\)

Since LHS = RHS, P(3) is true.

This formula is a well-known result and can be formally proven to be true for all natural numbers n using the Principle of Mathematical Induction. Thus, P(n): \(1^2 + 2^2 + ... + n^2 = \frac{n(n+1)(2n+1)}{6}\) is a statement true for all natural numbers n.

Solution: Let P(n) be the statement that \(4^n - 1\) is divisible by 3 for each natural number n.

Step I: Base Case (n=1)

For n = 1, the statement is \(P(1): 4^1 - 1 = 4 - 1 = 3\). Since 3 is divisible by 3, P(1) is true.

Step II: Inductive Hypothesis

Assume that P(k) is true for some arbitrary natural number k. That is, assume \(4^k - 1\) is divisible by 3.

This means we can write \(4^k - 1 = 3a\) for some integer a. Rearranging this, we get \(4^k = 3a + 1\).

Step III: Inductive Step (n=k+1)

We need to prove that P(k+1) is true, i.e., \(4^{k+1} - 1\) is divisible by 3.

Consider the expression for P(k+1):

$$4^\{k+1\}\; -\; 1\; =\; 4^k\; \backslash cdot\; 4\; -\; 1$$

Now, substitute \(4^k = 3a + 1\) from the inductive hypothesis:

$$=\; (3a\; +\; 1)\; \backslash cdot\; 4\; -\; 1$$

$$=\; 12a\; +\; 4\; -\; 1$$

$$=\; 12a\; +\; 3$$

$$=\; 3(4a\; +\; 1)$$

Since \(4a + 1\) is an integer, the expression \(3(4a + 1)\) is divisible by 3. Thus, \(4^{k+1} - 1\) is divisible by 3.

Conclusion:

Since P(1) is true and P(k) implies P(k+1), by the Principle of Mathematical Induction, the statement P(n): \(4^n - 1\) is divisible by 3 is true for all natural numbers n.

Solution: Let P(n) be the statement that \(2^{3n} - 1\) is divisible by 7 for all natural numbers n.

Step I: Base Case (n=1)

For n = 1, the statement is \(P(1): 2^{3\times 1} - 1 = 2^3 - 1 = 8 - 1 = 7\). Since 7 is divisible by 7, P(1) is true.

Step II: Inductive Hypothesis

Assume that P(k) is true for some arbitrary natural number k. That is, assume \(2^{3k} - 1\) is divisible by 7.

This means we can write \(2^{3k} - 1 = 7a\) for some integer a. Rearranging this, we get \(2^{3k} = 7a + 1\).

Step III: Inductive Step (n=k+1)

We need to prove that P(k+1) is true, i.e., \(2^{3(k+1)} - 1\) is divisible by 7.

Consider the expression for P(k+1):

$$2^\{3(k+1)\}\; -\; 1\; =\; 2^\{3k+3\}\; -\; 1\; =\; 2^\{3k\}\; \backslash cdot\; 2^3\; -\; 1$$

We know that \(2^3 = 8\). So, the expression becomes:

$$=\; 2^\{3k\}\; \backslash cdot\; 8\; -\; 1$$

Now, substitute \(2^{3k} = 7a + 1\) from the inductive hypothesis:

$$=\; (7a\; +\; 1)\; \backslash cdot\; 8\; -\; 1$$

$$=\; 56a\; +\; 8\; -\; 1$$

$$=\; 56a\; +\; 7$$

$$=\; 7(8a\; +\; 1)$$

Since \(8a + 1\) is an integer, the expression \(7(8a + 1)\) is divisible by 7. Thus, \(2^{3(k+1)} - 1\) is divisible by 7.

Conclusion:

Since P(1) is true and P(k) implies P(k+1), by the Principle of Mathematical Induction, the statement P(n): \(2^{3n} - 1\) is divisible by 7 is true for all natural numbers n.