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 $2^(m+1) > m+1$
$2^{m+1}=2 \times 2^m$
$2^{m+1}=2 \times 2^{m} > 2 \times m$
2m > m+1 ⇒ m > 1
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:$4+8+12+ \ldots +4k+4(k+1)=2k(k+1)((k+1)+1)$
$4+8+12+ \ldots +4k+4(k+1)=2k(k+1)((k+1)+1)$
$=(k+1)(2k+4)$
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
$1+2+3+\ldots+n<(2 n+1)^{2}$
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