CBSE Class 11 Mathematics Principle Of Mathematical Induction Worksheet Set 06

CBSE Class 11 Mathematics Worksheet - Principle of Mathematical Induction (5). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question 8. Show using PMI, \( (2n + 1) < (n + 3)^2 \)
Answer:
Let the given statement be denoted as \( P(n): (2n + 1) < (n + 3)^2 \).
First, we verify the base case for \( n = 1 \):
\( P(1): (2(1) + 1) < (1 + 3)^2 \)

\( \implies 3 < 16 \)
Since \( 3 \) is strictly less than \( 16 \), the base case \( P(1) \) holds true.

Next, we assume that the assertion \( P(k) \) is true for some positive integer \( k \):
\( P(k): 2k + 1 < (k + 3)^2 \)

Now, our objective is to establish that \( P(k + 1) \) is valid as well, which means we must show:
\( P(k + 1): 2k + 3 < (k + 4)^2 \)
or alternatively, \( 2k + 3 < k^2 + 8k + 16 \).

Starting with our assumption:
\( 2k + 1 < (k + 3)^2 \)
Let us add \( 2 \) to both sides of this inequality:

\( \implies (2k + 1) + 2 < (k + 3)^2 + 2 \)

\( \implies 2k + 3 < k^2 + 6k + 9 + 2 \)

\( \implies 2k + 3 < k^2 + 6k + 11 \)

For any natural number \( k \), we know that \( 6k + 11 < 8k + 16 \). Therefore, we can write:

\( \implies 2k + 3 < k^2 + 6k + 11 < k^2 + 8k + 16 \)

\( \implies 2k + 3 < k^2 + 8k + 16 \)

\( \implies 2k + 3 < (k + 4)^2 \)

This confirms that the statement \( P(k + 1) \) is indeed true.
Consequently, by the principle of mathematical induction, \( P(n) \) is valid for all \( n \in N \).
In simple words: To prove this inequality, we first show it works for \( n = 1 \). Then, assuming it holds for any number \( k \), we show that adding \( 2 \) to both sides keeps the inequality true for the next step, \( k + 1 \).

Exam Tip: Always explicitly write down the step where you compare the intermediate quadratic expression to the target expression. Stating the algebraic reason (like \( 6k + 11 < 8k + 16 \)) is crucial for scoring full marks.

Question 9. Show that \( 1^2 + 2^2 + \dots + n^2 > \frac{n^3}{3} \).
Answer:
Let us define the given statement as \( P(n): 1^2 + 2^2 + \dots + n^2 > \frac{n^3}{3} \).
First, we test the inequality for \( n = 1 \):
\( P(1): 1^2 > \frac{1^3}{3} \)

\( \implies 1 > \frac{1}{3} \)
This statement is obviously true, so the base case \( P(1) \) is verified.

Now, let us assume that the statement is correct for \( n = k \):
\( P(k): 1^2 + 2^2 + \dots + k^2 > \frac{k^3}{3} \)

We now need to prove that \( P(k+1) \) is also true, which means:
\( P(k+1): 1^2 + 2^2 + \dots + k^2 + (k+1)^2 > \frac{(k+1)^3}{3} \)
or, expanded as:
\( P(k+1): 1^2 + 2^2 + \dots + k^2 + (k+1)^2 > \frac{k^3 + 3k^2 + 3k + 1}{3} \)

Using our inductive hypothesis:
\( 1^2 + 2^2 + \dots + k^2 > \frac{k^3}{3} \)
Let us add \( (k+1)^2 \) to both sides of this relation:

\( \implies 1^2 + 2^2 + \dots + k^2 + (k+1)^2 > \frac{k^3}{3} + (k+1)^2 \)

\( \implies 1^2 + 2^2 + \dots + k^2 + (k+1)^2 > \frac{k^3 + 3(k^2 + 2k + 1)}{3} \)

\( \implies 1^2 + 2^2 + \dots + k^2 + (k+1)^2 > \frac{k^3 + 3k^2 + 6k + 3}{3} \)

We know that for any positive integer \( k \), the term \( 6k + 3 \) is strictly greater than \( 3k + 1 \). Therefore, it follows that:
\( \frac{k^3 + 3k^2 + 6k + 3}{3} > \frac{k^3 + 3k^2 + 3k + 1}{3} \)
Combining these inequalities gives us:

\( \implies 1^2 + 2^2 + \dots + k^2 + (k+1)^2 > \frac{k^3 + 3k^2 + 3k + 1}{3} \)

\( \implies 1^2 + 2^2 + \dots + k^2 + (k+1)^2 > \frac{(k+1)^3}{3} \)

Hence, the statement holds for \( P(k+1) \).
By the principle of mathematical induction, the statement is valid for all natural numbers \( n \in N \).
In simple words: To show that the sum of squares is larger than \( \frac{n^3}{3} \), we first verify it for \( 1 \). Then we assume it works for \( k \), add the next square \( (k+1)^2 \), and show the result is still larger than the target formula for \( k+1 \).

Exam Tip: Pay careful attention to the algebraic expansion of \( (k+1)^3 \). Showing the step where you demonstrate \( 6k + 3 > 3k + 1 \) clearly justifies the inequality shift to the examiner.

Question 10. Show by PMI, \( 1 + 2 + 3 + \dots + n < \frac{1}{8}(2n + 1)^2 \)

Answer:
Let us denote the given inequality as \( P(n): 1 + 2 + 3 + \dots + n < \frac{1}{8}(2n + 1)^2 \).
First, we verify the statement for \( n = 1 \):
\( P(1): 1 < \frac{1}{8}(2(1) + 1)^2 \)

\( \implies 1 < \frac{9}{8} \)
Since \( 1 < 1.125 \), the base case \( P(1) \) is clearly correct.

Now, we assume that the statement is true for \( n = k \):
\( P(k): 1 + 2 + 3 + \dots + k < \frac{1}{8}(2k + 1)^2 \)

We need to show that \( P(k+1) \) also holds true:
\( P(k+1): 1 + 2 + 3 + \dots + k + (k+1) < \frac{1}{8}(2k + 3)^2 \)
Expanding the right-hand side, this can be written as:
\( P(k+1): 1 + 2 + 3 + \dots + k + (k+1) < \frac{4k^2 + 12k + 9}{8} \)

By adding \( (k+1) \) to both sides of our assumed inequality, we get:

\( \implies 1 + 2 + 3 + \dots + k + (k+1) < \frac{1}{8}(2k + 1)^2 + (k+1) \)

\( \implies 1 + 2 + 3 + \dots + k + (k+1) < \frac{4k^2 + 4k + 1}{8} + (k+1) \)

\( \implies 1 + 2 + 3 + \dots + k + (k+1) < \frac{4k^2 + 4k + 1 + 8k + 8}{8} \)

\( \implies 1 + 2 + 3 + \dots + k + (k+1) < \frac{4k^2 + 12k + 9}{8} \)

\( \implies 1 + 2 + 3 + \dots + k + (k+1) < \frac{(2k + 3)^2}{8} \)

Thus, the statement is proved for \( P(k+1) \).
By PMI, \( P(n) \) is true for all natural numbers \( n \in N \).
In simple words: We prove this by showing it works for the first step, \( n = 1 \). Then, assuming the sum of numbers up to \( k \) is smaller than our formula, we add the next number \( k+1 \) and show the new sum is still smaller than the formula for \( k+1 \).

Exam Tip: Be careful when taking the LCM after adding \( (k+1) \). Expanding \( (2k+1)^2 \) to \( 4k^2 + 4k + 1 \) and properly grouping terms into the perfect square \( (2k+3)^2 \) is key to a smooth proof.

Question 11. Prove that \( (1 + x)^n \geq (1 + nx) \) for all natural no. \( n \text{, where } x > -1 \).
Answer:
Let the given statement be defined as \( P(n): (1 + x)^n \geq (1 + nx) \).
First, we verify the statement for the base case \( n = 1 \):
\( P(1): (1 + x)^1 \geq (1 + x) \)
Since both sides are equal, \( P(1) \) is clearly true.

Next, we assume that the statement holds for \( n = k \):
\( P(k): (1 + x)^k \geq (1 + kx) \)

We now aim to prove that \( P(k+1) \) is also valid, which means:
\( P(k+1): (1 + x)^{k+1} \geq 1 + (k+1)x \)
which can also be written as:
\( P(k+1): (1 + x)^{k+1} \geq 1 + kx + x \)

Given our assumption:
\( (1 + x)^k \geq 1 + kx \)
Since \( x > -1 \), the term \( (1 + x) \) is positive. Multiplying both sides of the inequality by \( (1 + x) \) preserves its direction:

\( \implies (1 + x)^k(1 + x) \geq (1 + kx)(1 + x) \)

\( \implies (1 + x)^{k+1} \geq 1 + x + kx + kx^2 \)

\( \implies (1 + x)^{k+1} \geq 1 + (k+1)x + kx^2 \)

Since \( k \) is a natural number and \( x^2 \) is always non-negative, the term \( kx^2 \) must be greater than or equal to \( 0 \). Hence:
\( 1 + (k+1)x + kx^2 \geq 1 + (k+1)x \)
Combining these inequalities, we get:

\( \implies (1 + x)^{k+1} \geq 1 + (k+1)x \)

Thus, the statement is valid for \( P(k+1) \).
By the principle of mathematical induction, the statement holds true for all natural numbers \( n \in N \).
In simple words: To show that \( (1+x)^n \) is at least \( 1+nx \), we first check it for \( n = 1 \). Then we assume it works for \( k \), multiply by \( (1+x) \), and use the fact that the leftover term \( kx^2 \) is always positive to show the inequality remains true for \( k+1 \).

Exam Tip: Explicitly mention that multiplying by \( (1+x) \) is allowed because \( x > -1 \) ensures that \( (1+x) \) is strictly positive. Failing to mention this constraint can result in lost marks.

Question 12. Prove that \( 2^n > n \) for all \( n \in N \).
Answer:
Let us define the given statement as \( P(n): 2^n > n \).
First, we test the assertion for the base case \( n = 1 \):
\( P(1): 2^1 > 1 \)
Since \( 2 \) is strictly greater than \( 1 \), the base case \( P(1) \) is correct.

Next, we assume that the statement is true for \( n = k \):
\( P(k): 2^k > k \)

We must now prove that the statement is valid for \( n = k + 1 \):
\( P(k+1): 2^{k+1} > k + 1 \)

Starting with our induction assumption:
\( 2^k > k \)
Multiplying both sides of the inequality by \( 2 \):

\( \implies 2^k \cdot 2 > 2k \)

\( \implies 2^{k+1} > k + k \)

Since \( k \) is a natural number, we have \( k \geq 1 \). This allows us to write:
\( k + k \geq k + 1 \)
Combining these inequalities gives:

\( \implies 2^{k+1} > k + 1 \)

This shows that \( P(k+1) \) is indeed true.
Consequently, by the principle of mathematical induction, the statement is valid for all natural numbers \( n \in N \).
In simple words: We prove this by showing it holds for the first number, \( 1 \). Then, assuming \( 2^k \) is larger than \( k \), we double both sides to show that the next power of two is larger than \( 2k \), which is naturally at least \( k+1 \).

Exam Tip: Don't skip the step justifying why \( k + k \geq k + 1 \). Since \( k \) is a natural number, \( k \geq 1 \) is a vital condition that makes this inequality leap logically valid to graders.