KC Sinha Mathematics Solution Class 11 Chapter 10 Principle of Mathematical Induction Exercise 10.1

 Exercise 10.1

Page no -10.16

Type 1

Question 1

Let $P(n)$ be the statement ' $1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\dfrac{n(n+1)(2 n+1)}{6}$. Show that $P(1), P(2)$ and $P(3)$ are true.

Sol :

To show that P(1), P(2) and P(3) are true, we’ll evaluate each statement individually.

Given:

The statement P(n) is 

$1^2+2^2+3^2+\ldots +n^2=\dfrac{n(n+1)(2n+1)}{6}$

Step 1: Verify P(1)

For n=1:

$1^2=\dfrac{1 \times (1+1) \times (2\times 1+1)}{6}=\dfrac{1\times 2 \times 3}{6}=\dfrac{6}{6}=1$

So, P(1) is true.

Step 2: Verify P(2)

For n=2

$1^2+2^2=1+4=5$

The formula gives:

$\dfrac{2 \times (2+1) \times (2 \times 2 +1)}{6}=\dfrac{2 \times 3 \times 5}{6}=\dfrac{3}{6}=5$

So, P(2) is true.

Step 3: Verify P(3) 

For n=3

$1^2+2^2+3^2=1+4+9=14$

The formula gives

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

So, P(3) is true

Therefore, we have verified that P(1), P(2) and P(3) are true.

Question 2

If $P(n)$ be the statement '10nt3 is a prime number', then prove that $P(1)$ $P(2)$ are true but $P(3)$ is false.

Sol :

Let's examine each statement P(n), where P(n) is defined as: "10n+3 is a prime number."

Step 1: Verify P(1) 

For n=1:

$10 \times 1 + 3=10+3=13$

Since 13 is a prime number, P(1) is true.

Step 2 : Verify P(2)

For n=2:

$10 \times 2 +3 =20+3=23$

Since 23 is a prime number,  P(2) is true.

Step 3 : Verify P(3)

For n=3:

$10 \times 3 +3 =30+3=33$

Since 33 is not a prime number (it is divided by 3 and 11), P(3) is false.

Thus, we have show that P(1), P(2) are true, but P(3) is false

Question 3

If $P(n)$ be the statement ' $2^{n}>n$ ' and if $P(m)$ is true, show that $P(m+1)$ also true.

Sol :

To show that if P(m) is true, then P(m+1) is also true, we'll assume that P(m) holds and then demonstrate that implies P(m+1) holds.

Given:

The statement P(n) is

$2^n > n$

Step 1: Assume P(m) is true

Assume that P(m) is true for some integer m , so

$2^m> m$

Step 2: Show that P(m+1) is true

We need to show that $2^(m+1) > m+1$

Starting with $2^{m+1}$, we can write it as:

$2^{m+1}=2 \times 2^m$

Since we assumed $2^m > m$, we can substitute this inequality in:

$2^{m+1}=2 \times 2^{m} > 2 \times m$

Now, if we can show that $2 \times m \geq m+1$ for sufficiently large m, this will imply $2^{m+1} > m+1$

2m > m+1 ⇒ m > 1

Thus, for $m \geq 2$, P(m+1) holds if P(m) holds.
This completes the proof by induction for $m \geq 2$

Prove the following by using the principle of mathematical induction :

Question 4

$4+8+12+\ldots+4 n=2 n(n+1)$ for all $n \in N$.

Sol :

Step 1: Base Case (n=1)

For n=1, the left-hand side of the statement is just the first term:

$4=2 \times 1 \times (1+1)=2 \times 1 \times 2=4$

So, the base case n=1 is true

Step 2: Inductive hypothesis

Assume that the statement is true for some integer n=k; that is assume:

$4+8+12+ \ldots +4k=2k(k+1)$

Step 3: Inductive step

We need to show that the statement is true for n=k+1; that is, we want to prove:

$4+8+12+ \ldots +4k+4(k+1)=2k(k+1)((k+1)+1)$

Using the inductive hypothesis, we can substitute $4+8+12+ \ldots +4k=2k(k+1)$

$4+8+12+ \ldots +4k+4(k+1)=2k(k+1)((k+1)+1)$

Now, factor out (k+1) on the right-hand side:

$=(k+1)(2k+4)$

Simplify inside the parentheses:

$(k+1)(2(k+2))=2(k+1)(k+2)$

This matches the right-hand side of the statement for n=k+1 , so the inductive step is complete

Conclusion

By the principle of mathematical induction, the statement

$4+8+12+\ldots +4n=2n(n+1)$

is true for all n∈N

Question 5

$1+2+3+\ldots+n=\dfrac{n(n+1)}{2}$

Sol :

Step 1: Base case (n=1)

For n=1, the left-hand side of the statement is just the first term:

$1=\dfrac{1 \times (1+1)}{2}=\dfrac{1 \times 2}{2}=1$

So, the base case n=1 is true

Step 2: Inductive hypothesis

Assume that the statement is true for some integer n=k; that is, assume:

$1+2+3+\ldots+k=\dfrac{k(k+1)}{2}$

Step 3: Inductive step

We need to show that the statement is true for n=k+1; that is, we want to prove:

$1+2+3+\ldots +k+(k+1)=\dfrac{(k+1)(k+1)+1}{2}$

Using the inductive hypothesis, we can write:

$1+2+3+\ldots +k+(k+1)=\dfrac{(k+1)}{2}+(k+1)$

Now, combine terms by factoring out (k+1):

$=\dfrac{k(k+1)+2(k+1)}{2}=\dfrac{(k+1)(k+2)}{2}$

This matches the right-hand side of the statement for n=k+1, so the inductive step is complete.

Conclusion

By the principle of mathematical induction, the statement

$1+2+3+\ldots +n=\dfrac{n(n+1)}{2}$

is true for all n∈N.

Question 6

$1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\dfrac{n(n+1)(2 n+1)}{6}$

Sol :

Step 1: Base case (n=1)

For n=1, the left-hand side of the statement is just the first term:

$1^2=1$

The right-hand side is:

$\dfrac{1(1+1) \times (2 \times 1 +1)}{6}=\dfrac{1 \times 2 \times 3}{6}=\dfrac{6}{6}=1$

So, the base case n=1 is true.

Step 2: Inductive Hypothesis

Assume that the statement is true for some integer n=k; that is, assume:

$1^{2}+2^{2}+3^{2}+\ldots+k^{2}=\dfrac{k(k+1)(2k+1)}{6}$

Step 3: Inductive step

We need to show that the statement is true for n=k+1; that is, we want to prove:

$1^{2}+2^{2}+3^{2}+\ldots+k^{2}+(k+1)^2=\dfrac{k(k+1)(k+2)(2k+1)+1}{6}$

Using the inductive hypothesis, we can write:

$1^{2}+2^{2}+3^{2}+\ldots+k^{2}+(k+1)^2=\dfrac{k(k+1)(2k+1)}{6}+(k+1)^2$

Now, factor out (k+1) on the right-hand side:

$=\dfrac{k(k+1)(2k+1)+6(k+1)^2}{6}$

Factor (k+1) from both terms in the numerator:

$=\dfrac{(k+1)(k(2k+1)+6(k+1))}{6}$

Simplify inside the parentheses:

$=\dfrac{(k+1)(2k^2+k+6k+6)}{6}=\dfrac{(k+1)(2k^2+7k+6)}{6}$

Now, factor $2k^2+7k+6$ as (k+2)(2k+1): $=\dfrac{(k+1)(k+2)(2k+1)}{6}$

Question 7

$1+3+3^{2}+\ldots+3^{n-1}=\dfrac{3^{n}-1}{2}$

Sol :

Step 1: Base case (n=1)

For n=1, the left-hand side of the statement is just the first term:

The right-hand side for n=1 is:

$\dfrac{3^1-1}{2}=\dfrac{3-1}{2}=\dfrac{2}{2}=1$

So, the base case n=1 is true.

Step 2: Inductive Hypothesis

Assume that the statement is true for some integer n=k; that is, assume:

$1+3+3^2+ \ldots +3^{k-1}=\dfrac{3^k-1}{2}$

Step 3: Inductive Step

We need to show that the statement is true for n=k+1; that is, we want to prove:

$1+3+3^2+ \ldots +3^{k-1}+3^k=\dfrac{3^{k+1}-1}{2}$

Using the inductive hypothesis, we can write:

$1+3+3^2+ \ldots +3^{k-1}+3^k=\dfrac{3^{k}-1}{2}+3^k$

Now, combine the terms on the right-hand side by finding a common denominator:

$=\dfrac{3^k-1+2 \times 3^k}{2}=\dfrac{3^k+2 \times 3^k -1}{2}$

Combine the $3^k$ terms:

$=\dfrac{3 \times 3^k-1}{2}=\dfrac{3^{k+1}-1}{2}$

This matches the right-hand side of the statement for n=k+1, so the inductive step is complete.

Conclusion

By the principle of mathematical induction, the statement

$1+3+3^2+ \ldots +3^{n-1}=\dfrac{3^n-1}{2}$

is true for all n∈N.

Question 8

$\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\ldots+\dfrac{1}{2^{n}}=1-\dfrac{1}{2^{n}}$

Sol :

Step 1: Base Case (n=1)

For n=1, the left-hand side of the statement is just the first term: $\dfrac{1}{2}$

The right-hand side for n=1 is:

$1-\dfrac{1}{2}=\dfrac{2-1}{2}=\dfrac{1}{2}$

So, the base case n=1 is true.

Step 2: Inductive Hypothesis

Assume that the statement is true for some integer n=k; that is, assume:

$\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+ \ldots \dfrac{1}{2^k}=1-\dfrac{1}{2^k}$

Step 3: Inductive Step

We need to show that the statement is true for n=k+1; that is, we want to prove:

$\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+ \ldots \dfrac{1}{2^k}+\dfrac{1}{2^{k+1}}=1-\dfrac{1}{2^{k+1}}$

Using the inductive hypothesis, we can write:

$\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+ \ldots \dfrac{1}{2^k}+\dfrac{1}{2^{k+1}}=\left( 1-\dfrac{1}{2^k} \right)+\dfrac{1}{2^{k+1}}$

Now, combine the terms on the right-hand side:

$=1-\dfrac{1}{2^k}+\dfrac{1}{2^{k+1}}$

To simplify, rewrite $\dfrac{1}{2^k}$ as $\dfrac{2}{2^{k+1}}$:

$1-\dfrac{2}{2^{k+1}}+\dfrac{1}{2^{k+1}}=1-\dfrac{2-1}{2^{k+1}}=1-\dfrac{1}{2^{k+1}}$

This matches the right-hand side of the statement for n=k+1, so the inductive step is complete.

Conclusion

By the principle of mathematical induction, the statement

$\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+ \ldots +\dfrac{1}{2^n}=1-\dfrac{1}{2^n}$

is true for all n∈N.

Question 9

$(a b)^{n}=a^{n} b^{n}$

Sol :

Step 1: Base case (n=1)

Foe n=1, the left-hand side of the statement is:

$(ab)^1=ab$

The right-hand side is:

$a^1b^1=ab$

So, the base case n=1 is true

Step 2: Inductive Hypothesis

Assume that the statement is true for some integer n=k; that is, assume:

$(ab)^k=a^kb^k$

Step 3: Inductive step

We need to show that the statement is true for n=k+1; that is, we want to prove:

$(ab)^{k+1}=a^{k+1}b^{k+1}$

Using the inductive hypothesis, we can write:

$(ab)^{k+1}=(ab)^k \times (ab)$

By the inductive hypothesis, we know that $(ab)^k=a^kb^k$, so we can substitute this in:

$=a^k b^k \times ab$

Now, use the properties of exponents to combine terms:

$=a^{k+1}b^{k+1}$

This matches the right-hand side of the statement for n=k+1, so the inductive step is complete.

Conclusion

By the principle of mathematical induction, the statement

$(ab)^n=a^n b^n$

is true for all n∈N.

Question 10

$1^{2}+3^{2}+5^{2}+\ldots+(2 n-1)^{2}=\dfrac{n(2 n-1)(2 n+1)}{3}$

Sol :

Step 1: Base case (n=1)

For n=1, the left-hand side of the statement is just the first term:

$1^2=1$

The right hand side for n=1 is:

$=\dfrac{1(2 \times 1 -1) (2 \times 1 +1)}{3}=\dfrac{1 \times 1 \times 3}{3}=\dfrac{3}{3}=1$

So, the base case n=1 is true

Step 2: Inductive Hypothesis

Assume that the statement is true for some integer n=k; that is, assume:

$1^{2}+3^{2}+5^{2}+\ldots+(2k-1)^{2}=\dfrac{k(2k-1)(2k+1)}{3}$

Step 3: Inductive step

We need to show that the statement is true for n=k+1; that is, we want to prove:

$1^2+3^2+5^2+ \ldots +(2k-1)^2+(2(k+1)-1)^2=\dfrac{(k+1)(2(k+1)-1)(2(k+1)+1}{3}$

Using the inductive hypothesis, we can write:

$1^2+3^2+5^2+ \ldots +(2k-1)^2+(2(k+1)-1)^2=\dfrac{k(2k-1)(2k+1)}{3}+(2k+1)^2$

Now, expand $(2k+1)^2$ and combine it with the term from the inductive hypothesis:

$=\dfrac{k(2k-1)(2k+1)+3(2k+1)^2}{3}$

Factor 2k+1 out of the numerator:

$=\dfrac{(2k+1)(k(2k-1)+3(2k+1))}{3}$

Now expand and simplify inside the parentheses:

$=\dfrac{(2k+1)(2k^2-k+6k+3)}{3}=\dfrac{(2k+1)(2k^2+5k+3)}{3}$

Now, rewrite 2k+1 and 2k+3 in terms of k+1:

$=\dfrac{(k+1)(2(k+1)-1)(2(k+1)+1)}{3}$

This is exactly the form we wanted to show for n=k+1:

$1^2+3^2+5^2+ \ldots +(2(k+1)-1)^2=\dfrac{(k+1)(2(k+1)+1)}{3}$

Conclusion

By the principle of mathematical induction, we have shown that

$1^{2}+3^{2}+5^{2}+\ldots+(2 n-1)^{2}=\dfrac{n(2 n-1)(2 n+1)}{3}$

is true for all n∈N

Question 11

$1^{3}+3^{3}+5^{3}+\ldots+(2 n-1)^{3}=n^{2}\left(2 n^{2}-1\right)$

Sol :

Step 1: Base Case (n=1)

For n=1, the left-hand side of the statement is just the first term:

$1^3=1$

The right-hand side for n=1 is:
$1^2(2 \times 1^2 -1)=1 \times (2-1)=1$

So, the base case n=1 is true.

Step 2: Inductive Hypothesis

Assume that the statement is true for some integer n=k; that is, assume:

$1^3+3^3+5^3+ \ldots +(2k-1)^3=k^2 (2k^2-1)$

Step 3: Inductive Step

We need to show that the statement is true for n=k+1; that is, we want to prove:

$1^3+3^3+5^3+ \ldots +(2k-1)^3+(2(k+1)-1)^3=(k+1)^2(2(k+1)^2-1)$

Using the inductive hypothesis, we can write:

$1^3+3^3+5^3+ \ldots +(2k-1)^3+(2(k+1)-1)^3=k^2(2k^2-1)+(2k+1)^3$

Expanding $(2k+1)^3$

Now, calculate $(2k+1)^3$:

$(2k+1)^3=8k^3+12k^2+6k+1$

Substituting and Simplifying

Substitute back to get:

$k^2(2k^2-1)+(2k+1)^3=k^2(2k^2-1)+8k^3+12k^2+6k+1$

Now let's expand $k^2(2k^2-1)$ as follows:

$k^2(2k^2-1)=2k^4-k^2$

Substituting this back in, we have:

$2k^4-k^2+8k^3+12k^2+6k+1$

Let's combine like terms:

$=2k^4+8k^3+(12k^2-k^2)+6k+1$

$=2k^4+8k^3+11k^2+6k+1$

we need to show that the expression we've obtained, matches the right-hand side of the induction hypothesis for n=k+1, which is

$(k+1)^2 (2(k+1)^2-1)$

Step 4: Expand $(k+1)^2(2(k+1)^2-1)$

1. Expand $(k+1)^2$:

 $(k+1)^2=k^2+2k+1$

2. Expand $2(k+1)^2-1:$

 $2(k+1)^2-1=2(k^2+2k+1)-1=2k^2+4k+2-1=2k^2+4k+1$

3.Combine them:

$(k+1)^2 (2(k+1)^2-1)=(k^2+2k+1)(2k^2+4k+1)$

Step 5: Distribute the terms

Now, let's distribute:

$(k^2+2k+1)(2k^2+4k+1)=k^2(2k^2+4k+1)+2k(2k^2+4k+1)+1(2k^2+4k+1)$

Calculate each part:

1. For $k^2(2k^2+4k+1):$

$=2k^4+4k^3+k^2$

2. For $2k(2k^2+4k+1)$:

$=4k^3+8k^2+2k$

3. For $1(2k^2+4k+1)$:

$=2k^2+4k+1$

Step 6: Combine all the terms

Now let's combine all these results:

$2k^4+4k^3+k^2+4k^3+8k^2+2k+2k^2+4k+1$

Combining like terms:

$k^4$ term: $2k^4$

$k^3$ term: $4k^3+4k^3=8k^3$

$k^2$ term: $k^2+8k^2+2k^2=11k^2$

k terms: 2k+4=6k

constant term: 1

Putting it all together, we get:

$2k^4+8k^3+11k^2+6k+1$

Conclusion:

Thus, we have shown:

$1^3+3^3+5^3+ \ldots + (2(k+1)-1)^3=(k+1)^2 (2(k+1)^2-1)$

Since the base case holds and the inductive step has been proven, by the principle of mathematical induction, we conclude that:

$1^3+3^3+5^3+ \ldots +(2n-1)^3=n^2(2n^2-1)$

is true for all natural numbers n

Question 12

$3 \cdot 2^{2}+3^{2} \cdot 2^{3}+\ldots+3^{n} \cdot 2^{n+1}=\dfrac{12}{5}\left(6^{n}-1\right)$

Sol :

Step 1: Base Case (n = 1)

For n=1:

Left-hand side:

$3 \times 2^2 = 3 \times 4=12$

Right-hand side:

$\dfrac{12}{5}(6^1-1)=\dfrac{12}{5}(6-1)=\dfrac{12}{5} \times 5=12$

Since both sides are equal, the base case holds true.

Step 2: Inductive Hypothesis

Assume the statement is true for some integer n=k

$3 \times 2^2 + 3^2 \times 2^3 + \ldots + 3^k \times 2^{k+1}=\dfrac{12}{5}(6^k-1)$

Step 3: Inductive Step

We need to show that the statement is true for n=k+1:

$3 \times 2^2 +3^2 \times 2^3 + \ldots + 3^k \times 2^{k+1}+3^{k+1} \times 2^{k+2}=\dfrac{12}{5}(6^{k+1}-1)$

Using the inductive hypothesis, we have:

$\dfrac{12}{5}(6^k-1)+3^{k+1} \times 2^{k+2}$

Now, let's simplify this expression.

Step 4: Combine Terms

Starting with: 

$\dfrac{12}{5}(6^k-1)+3^{k+1} \times 2^{k+2}$

We can rewrite $3^{k+1} \times 2^{k+2}$ to match the form we need. Note that $2^{k+2}=2 \times 2^{k+1}$

$=\dfrac{12}{5}(6^k-1)+3^{k+1} \times 2 \times 2^{k+1}$

Step 5: Rewrite $3^{k+1} \times 2^{k+2}$

Notice that

$3^{k+1} \times 2^{k+2}=2 \times 3^{k+1} \times 2^{k+1}$

So the equation becomes:

$=\dfrac{12}{5}(6^k-1)+2 \times  3^{k+1} \times 2^{k+1}$

Step 6: Factor and Expand

Now we want to express this in terms of $6^{k+1}$. Recall that $6^{k+1}=6 \times 6^{k}$, we have:

$=\dfrac{12}{5}(6^k)+\dfrac{12}{5}(-1)+2 \times 3^{k+1} \times 2^{k+1}$

$=\dfrac{12 \times 6^k}{5}-\dfrac{12}{5}+2 \times 3^{k+1} \times 2^{k+1}$

Step 7: Final Formulation

Rearranging gives : $=\dfrac{12 \times 6^k -12 +12 \times 6^k}{5}$

This simplifies to:

$=\dfrac{12 \times 6^{k+1}-12}{5}$

Conclusion

Thus we have:

$3 \times 2^2 +3^2 \times 2^3 + \ldots + 3^{k+1} \times 2^{k+2}=\dfrac{12}{5}(6^{k+1}-1)$

By the principle of mathematical induction, the statement is true for all natural numbers  n

Question 13

$1.3+2 \cdot 3^{2}+3 \cdot 3^{3}+\ldots+n \cdot 3^{n}=\dfrac{(2 n-1) 3^{n+1}+3}{4}$

Sol :

Step 1: Base case (n=1)

For n=1

Left hand side: $1 \times 3 =3$

Right hand side:

$\dfrac{(2 \times 1 -1) \times 3^{1+1}+3}{4}=\dfrac{(2-1) \times 3^2 +3}{4}=\dfrac{1 \times 9 +3}{4}=\dfrac{12}{4}=3$

Since both sides are equal, the base case holds true.

Step 2: Inductive Hypothesis

Assume the statement is true for some integer n=k

$1\times 3 +2 \times 3^2+ 3\times 3^3 + \ldots +k \times 3^k=\dfrac{(2k-1) \times 3^{k+1}+3}{4}$

Step 3:  Inductive Step

We need to show that the statement is true for n=k+1; that is, we want to prove:

$1 \times 3 + 2\times 3^2 +3 \times 3^2 + \ldots + k \times 3^k + (k+1) \times 3^{k+1} \times 3^{k+1}=\dfrac{(2(k+1)-1) \times 3^{(k+1)+1}+3}{4}$

Using the inductive hypothesis, we have:

$\dfrac{(2k-1) \times 3^{k+1}+3}{4}+(k+1) \times 3^{k+1}$

Step 4: Simplify the Expression

Factor out $3^{k+1}$ from the terms:

$=\dfrac{(2k-1) \times 3^{k+1}+3}{4}+\dfrac{4(k+1) \times 3^{k+1}}{4}$

Combine the terms over a common denominator:

$=\dfrac{((2k-1)+4(k+1))\times 3^{k+1}+3}{4}$

Expand inside the parentheses:

$=\dfrac{(2k-1+4k+4) \times 3^{k+1}+3}{4}$

Combine like terms:

$=\dfrac{(6k+3) \times 3^{k+1}+3}{4}$

Factor out 3:

$=\dfrac{3(2k+1) \times 3^{k+1}+3}{4}$

Rewrite this as:

$=\dfrac{(2(k+1)-1) \times 3^{(k+1)+1}+3}{4}$

Conclusion

By the principle of mathematical induction, the statement is true for all natural numbers n

 

Question 14

$a+a r+a r^{2}+\ldots+a r^{n-1}=\dfrac{a\left(1-r^{n}\right)}{1-r}, r \neq 1$

Sol :

Step 1: Base case (n=1)

For n=1

Left-hand side:

a

Right-hand side:

$=\dfrac{a(1-r^1)}{1-r}=\dfrac{a(1-r)}{1-r}=a$

Since both sides are equal, the base case holds.

Step 2: Inductive Hypothesis

Assume that the formula is true for some integer n=k; that is, assume:

$a+ar+ar^2+\ldots +ar^{k-1}=\dfrac{a(1-r^k)}{1-r}$

Step 3: Inductive Step

We need to show that the formula holds for n=k+1; that is, we want to prove:

$a+ar+ar^2+\ldots +ar^{k-1}+ar^k=\dfrac{a(1-r^{k+1})}{1-r}$

Using the inductive hypothesis, we can write:

$a+ar+ar^2+\ldots +ar^{k-1}+ar^k=\dfrac{a(1-r^k)}{1-r}+ar^k$

Step 4: Simplify the Expression

To combine the terms on the right-hand side, let's rewrite $ar^k$ as a fraction with the same denominator:

$=\dfrac{a(1-r^k)}{1-r}+\dfrac{ar^k(1-r)}{1-r}$

Expanding $ar^k(1-r)$ in the numerator:

$=\dfrac{a(1-r^k)+ar^k-ar^{k+1}}{1-r}$

Combine terms in the numerator:

$=\dfrac{a(1-r^{k+1})}{1-r}$

Conclusion

Since we have shown that the statement holds for n=k+1 if it holds for n=k, by the principle of mathematical induction, the formula is true for all natural numbers n when $r /neq 1$

Question 15

$a+(a+d)+(a+2 d)+\ldots+[a+(n-1) d]=\dfrac{n}{2}[2 a+(n-1) d]$

Sol :

Step 1: Base case (n=1)

For n=1:

Left-hand side:  a

Right-hand side: $\dfrac{1}{2} \left[ 2a+(1-1)d\right]=\dfrac{1}{2} \times 2a =a$

Since both sides are equal, the base case holds.

Step 2: Inductive Hypothesis

Assume that the formula is true for some integer n=k; that is, assume:

$a+(a+d)+(a+2d)+\ldots +[a+(k-1)d]=\dfrac{k}{2} \left[ 2a+ (k-1)d\right]$

Step 3: Inductive Step

We need to show that the formula holds for n=k+1; that is, we want to prove:

$a+(a+d)+(a+2d)+\ldots +[a+(k-1)d]+[a+kd]=\dfrac{k+1}{2} \left[ 2a+ k \times d\right]$

Using the inductive hypothesis, we can write:

$a+(a+d)+(a+2d)+\ldots +[a+(k-1)d]+[a+kd]=\dfrac{k}{2} \left[ 2a+(k-1)d\right] + [a+kd]$

Step 4: Simplify the Expression

Now, substitute a+kd and simplify:

$=\dfrac{k}{2} \left[ 2a+ (k-1)d \right]+a+kd$

we can write $a+kd$ as $\dfrac{2a+kd}{2}$ to combine terms with a common denominator:

$=\dfrac{k[2a+(k-1)d]+2a+2kd}{2}$

we have:

$=\dfrac{k[2a+(k-1)d]+2a+2kd}{2}$

Now, let's expand $k[2a+(k-1)d]$ and simplify the expression:

$=\dfrac{(2ak+k^2d-kd)+2a+2kd}{2}$

Combine terms in the numerator:

1. Combine terms with a: $2ak+2a=2a(k+1)$

2. Combine terms with d: $k^2d-kd+2kd=d(k^2+k)$

Putting it all together:

$=\dfrac{2a(k+1)+d(k+1)k}{2}$

Now factor out (k+1):

$=\dfrac{(k+1)(2a+kd)}{2}$

This matches the right-hand side of our statement for n=k+1:

$=\dfrac{(k+1)}{d} \left[ 2a+kd\right]$

Conclusion

Since both sides are now equal, we have shown that if the formula holds for n=k, then it also holds for n=k+1. With the base case verified, the principle of mathematical induction confirms that:

$a+(a+d)+(a+2d)+ \ldots + [a+(n-1)d]=\dfrac{n}{2} \left[ 2a+(n-1)d\right]$

holds for all n∈N.

Question 16

$\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\ldots+\dfrac{1}{(2 n-1)(2 n+1)}=\dfrac{n}{2 n+1}$

Sol :

Step 1: Base case (n=1)

For n=1, the left-hand side has just one term:

$\dfrac{1}{1\times 3}=\dfrac{1}{3}$

The right-hand side for n=1 is:

$=\dfrac{1}{2 \times 1+1}=\dfrac{1}{3}$

Both sides are equal, so the base case holds.

Step 2: Inductive Hypothesis

Assume that the statement is true for n=k; that is, assume:

$\dfrac{1}{1 \times 3}+\dfrac{1}{3 \times 5}+\dfrac{1}{5 \times 7}+ \ldots + \dfrac{1}{(2k-1)(2k+1)}=\dfrac{k}{2k+1}$

Step 3: Inductive step

We need to show that the formula holds for n=k+1; that is , we want to prove:

$\dfrac{1}{1 \times 3}+\dfrac{1}{3 \times 5} +\dfrac{1}{5 \times 7} + \ldots + \dfrac{1}{(2k-1)(2k+1)}+\dfrac{1}{(2k+1)(2k+3)}=\dfrac{k+1}{2(k+1)+1}$

Using the inductive hypothesis, we can rewrite the left-hand side as:

$\dfrac{k}{2k+1}+\dfrac{1}{(2k+1)(2k+3)}$

Combine these terms over a common denominator (2k+1)(2k+3):

$=\dfrac{k(2k+3)+1}{(2k+1)(2k+3)}$

Now expand and simplify the numerator:

$=\dfrac{2k^2+3k+1}{(2k+1)(2k+3)}$

This expression can be factored as:

$=\dfrac{(k+1)(2k+1)}{(2k+1)(2k+3)}$

Cancel out (2k+1) in the numerator and denominator:

$=\dfrac{k+1}{2k+3}$

Conclusion:

We have shown that if the formula holds for n=k, then it also holds for n=k+1. Since the base case is also true, by the principle of mathematical induction, the formula

$\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\ldots+\dfrac{1}{(2 n-1)(2 n+1)}=\dfrac{n}{2 n+1}$

holds for all n∈N.

Question 17

$\dfrac{1}{3.7}+\dfrac{1}{7.11}+\dfrac{1}{11.15}+\ldots+\dfrac{1}{(4 n-1)(4 n+3)}=\dfrac{n}{3(4 n+3)}$

Question 18

$1.2+2.3+3.4+\ldots+n(n+1)=\dfrac{n(n+1)(n+2)}{3}$

Question 19

$1.3+3.5+5.7+\ldots+(2 n-1)(2 n+1)=\dfrac{n\left(4 n^{2}+6 n-1\right)}{3}$

Question 20

$\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\ldots+\dfrac{1}{(3 n-2)(3 n+1)}=\dfrac{n}{3 n+1}$

Page no -10.17

Question 21

$\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.1}+\ldots+\dfrac{1}{(3 n-1)(3 n+2)}=\dfrac{n}{6 n+4}$

Question 22

$\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\ldots+\dfrac{1}{(2 n+1)(2 n+3)}=\dfrac{n}{3(2 n+3)}$

Question 23

$\dfrac{1}{1.2 .3}+\dfrac{1}{2.3 .4}+\dfrac{1}{3.4 .5}+\ldots+\dfrac{1}{n(n+1)(n+2)}=\dfrac{n(n+3)}{4(n+1)(n+2)}$

Question 24

$1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\ldots+\dfrac{1}{1+2+3+\ldots+n}=\dfrac{2 n}{n+1}$

Question 25

$\left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right) \ldots\left(1+\dfrac{1}{n}\right)=n+1$

Question 26

$\left(1+\dfrac{3}{1}\right)\left(1+\dfrac{5}{4}\right)\left(1+\dfrac{7}{9}\right) \ldots\left(1+\dfrac{2 n+1}{n^{2}}\right)=(n+1)^{2}$

Question 27

$(\cos \theta+i \sin \theta)^{n}=\cos n \theta+i \sin n \theta$

Question 28

$3.6+6.9+9.12+\ldots+3 n(3 n+3)=3 n(n+1)(n+2)$

Question 29

If $u_{0}=2, u_{1}=3$ and $u_{n+1}=3 u_{n}-2 u_{n-1}$, show that $u_{n}=2^{n}+1, n \in N$.

Question 30

If $a_{0}=0, a_{1}=1$ and $a_{n+1}=3 a_{n}-2 a_{n-1}$, prove that $a_{n}=2^{n}-1$

Question 31

If $A_{1}=\cos \theta, A_{2}=\cos 2 \theta$ and for every natural number $m>2$, $A_{m}=2 A_{m-1} \cos \theta-A_{m-2}$, prove that $A_{n}=\cos n \theta$.

 Type 2

In $n$ is a natural number, using mathematical induction show that :

Question 32

32.$n(n+1)(n+5)$ is divisible by 6. 33. $n^{7}-n$ is a multiple of 7 .

Question 34

34.$4^{n}-3 n-1$ is divisible by 9

35. $9^{n}-8 n-1$ is divisible by 64 .

Question 36

$10^{2 n-1}+1$ is divisible by 11

Question 37

$x^{n}-y^{n}$ is divisible by $x+y$ when $n$ is an even integer.

Question 38

38.$2.7^{n}+3.5^{n}-5$ is divisible by $24

39. 3^{2 n+2}-8 n-9$ is divisible by 8

Question 40

40.$x^{2 n}-y^{2 n}$ is divisible by $x+y$

41. $7^{n}-3^{n}$ is divisible by 4 .

Question 42

$(41)^{n}-(14)^{n}$ is a multiple of 27

 Type 3

Show by using mathematical induction that: 

Question 43

(i) $2^{n}>n$

(ii) $2^{n}>n^{2}, n \geq 5, n \in N$.

Question 44

$1^{2}+2^{2}+\ldots+n^{2}>\dfrac{n^{3}}{3}$

Question 45

$\dfrac{1}{n+1}+\dfrac{1}{n+2}+\ldots+\dfrac{1}{2 n}>\dfrac{13}{24}, n>1, n \in N$

Question 46

$1+\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+\ldots+\dfrac{1}{n^{2}}<2-\dfrac{1}{n}, n \geq 2, n \in N$.

Question 47

$1+2+3+\ldots+n<(2 n+1)^{2}$