Exercise 4.4-Additional Questions - Chapter 4 Combinatorics and Mathematical Induction 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Additional Questions
Question 1.
Prove by induction the inequality $(1+x)^n \geq 1+n x$, whenever $x$ is positive and $n$ is a positive integer.
Solution:
P(n) : $(1+\mathrm{x})^{\mathrm{n}} \geq 1+\mathrm{nx}$
$\mathrm{P}(1):(1+\mathrm{x})^1 \geq 1+\mathrm{x}$
$\Rightarrow 1+x \geq 1+x$, which is true.
Hence, $\mathrm{P}(1)$ is true.
Let $\mathrm{P}(\mathrm{k})$ be true
(i.e.) $(1+x)^k \geq 1+k x$
We have to prove that $P(k+1)$ is true.
(i.e.) $(1+\mathrm{x})^{\mathrm{k}+1} \geq 1+(\mathrm{k}+1) \mathrm{x}$
Now, $(1+\mathrm{x})^{\mathrm{k}+1} \geq 1+\mathrm{kx}[\because \mathrm{p}(\mathrm{k})$ is true $]$
Multiplying both sides by $(1+x)$, we get
$
\begin{aligned}
& (1+x)^k(1+x) \geq(1+k x)(1+x) \\
& \Rightarrow(1+x)^{k+1} \geq 1+k x+x+k x^2 \\
& \Rightarrow(1+x)^{k+1} \geq 1+(\mathrm{k}+1) \mathrm{x}+\mathrm{kx}^2
\end{aligned}
$
Now, $1+(\mathrm{k}+1) \mathrm{x}+\mathrm{kx}^2 \geq 1+(\mathrm{k}+1) \mathrm{x}$
$
\left[\because \mathrm{kx}^2>0\right]
$
From (1) and (2), we get
$
(1+x)^{k+1} \geq 1+(k+1) x
$

$\therefore \mathrm{P}(\mathrm{k}+1)$ is true if $\mathrm{P}(\mathrm{k})$ is true.
Hence, by the principle of mathematical induction, $P(n)$ is true for all values, of $n$.
Question 2.
$3^{2 \mathrm{n}}-1$ is divisible by 8
Solution:
$\mathrm{P}(\mathrm{n})=3^{2 \mathrm{n}}-1$ is divisible by 8
For $n=1$, we get
$P(1)=3^{2.1}-1=9-1=8$
$P(1)=8$, which is divisible by 8 .
Let $P(n)$ be true for $n=k$
$P(k)=3^{2 \mathrm{k}}-1$ is divisible by 8
Now, $P(k+1)=3^{(2 k+2)}-1=3^{2 \mathrm{k}} \cdot 3^2-1$
$=3^2\left(3^{2 \mathrm{k}}-1\right)+8$
Now, $3^{2 \mathrm{k}}-1$ is divisible by 9 . [Using (1)]
$\therefore 3^2\left(3^{2 \mathrm{k}}-1\right)+8$ is also divisible by 8 .
Hence, $3^{2 \mathrm{n}}-1$ is divisible by $8 \forall \mathrm{n} E N$
Question 3.
Prove by the principle of mathematical induction if $\mathrm{x}$ and $\mathrm{y}$ are any two distinct integers, then $\mathrm{x}^{\mathrm{n}}-\mathrm{y}^{\mathrm{n}}$ is divisible by $\mathrm{x}-\mathrm{y}$. [OR]
$\mathrm{x}^{\mathrm{n}}-\mathrm{y}^{\mathrm{n}}$ is divisible by $\mathrm{x}-\mathrm{y}$, where $\mathrm{x}-\mathrm{y} \neq 0$.
Solution:
Let the given statement be $P(n)$.
(i.e.) $P(n): x^n-y^n=M(x-y), x-y \neq 0$

Step I.
When $\mathrm{n}=1$,
$x^n-y^n=x-y=M(x-y)$ $\Rightarrow P(1)$ is true.
Step II.
Assume that $\mathrm{P}(\mathrm{k})$ is true.
$
\text { (i.e.) } x^k-y^k=M(x-y), x-y \neq 0
$
We shall now show that $P(k+1)$ is true
$
\begin{aligned}
& \text { Now, } x^{k+1}-y^{k+1}=x^{k+1}-x^k y+x^{k+1} y-y^{k+1} \\
& =x^k(x-y)+y\left(x^k-y^k\right) \\
& =x^k(x-y)+y M(x-y)[U s n g ~ \ldots . .(1)] \\
& =(x-y)\left(x^k-y M\right)
\end{aligned}
$
$\therefore$ By the principle of mathematical induction, $\mathrm{P}(\mathrm{n})$ is true for all $\mathrm{n} \in \mathrm{N}$
Question 4.
Prove by the principle of mathematical induction that for every natural number $n, 3^{2 n+2}-8 n-9$ is divisible by 8 .
Solution:
Let $P(n): 3^{2 n+2}-8 n-9$ is divisible by 8 .

Then, $P(1): 3^{2.1+2}-8.1-9$ is divisible by 8 .
(i.e.) $3^4-8-9$ is divisible by 8 or $81-8-9$ is divisible by 8
(or) 64 is divisible by 8 , which is true.
Suppose $P(k)$ is true, then
$\mathrm{P}(\mathrm{k}): 3^{2 \mathrm{k}+2}-8 \mathrm{k}-9$ is divisible by 8
(i.e.) $3^{2 \mathrm{k}+2}-8 \mathrm{k}-9=8 \mathrm{~m}$, where $\mathrm{m} \in \mathrm{N}$ (or)
$3^{2 \mathrm{k}+2}=8 \mathrm{~m}+8 \mathrm{k}+9$
$\mathrm{P}(\mathrm{k}+1)$ is the statement given by, ...(1)
$
\mathrm{P}(\mathrm{k}+1): 3^{2(\mathrm{k}+1)+2}-8(\mathrm{k}+1)-9
$
Now, $3^{2(k+1)+2}-8(k+1)-9=3^{2 k+4}-8 k-8-9=3^{2 k+4}-8 k-17$
$
\begin{aligned}
& =9.3^{2 k+2}-8 k-17 \\
& =9(8 m+8 k+9)-8 k-17
\end{aligned}
$
$[\because \mathrm{P}(k)$ is true. Using $(1)]$
$\therefore 3^{2(k+1)+2}-8(k+1)-9$ is divisible by 8 .
$\therefore \mathrm{P}(\mathrm{k}+1)$ is true
Hence, by the principle of mathematical induction, $\mathrm{P}(\mathrm{n})$ is true for all $n \in N$

Question 5.
Use the principle of mathematical induction to prove that for every natural number $n$.
$
\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots\left(1+\frac{(2 n+1)}{n^2}\right)=(n+1)^2
$
Solution:
Let $\mathrm{P}(\mathrm{n})$ be the given statement, i.e.
$
\mathrm{P}(n):\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots\left(1+\frac{(2 n+1)}{n^2}\right)=(n+1)^2
$
For $n=1, \mathrm{LHS}=1+\frac{3}{1}=4$
and
$
\therefore \quad \text { LHS }=\text { RHS }
$
$\Rightarrow \mathrm{P}(1)$ is true.
We note than $P(n)$ is true for $n=1$.
Assume that $P(\mathrm{k})$ is true
$
\mathrm{P}(k):\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots\left(1+\frac{(2 k+1)}{k^2}\right)=(k+1)^2
$
Now, we shall prove that $P(k+1)$ is true whenever $P(k)$ is true. We have,

$
\begin{aligned}
\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right) & \cdots\left(1+\frac{(2 k+1)}{k^2}\right)\left(1+\frac{2(k+1)+1}{(k+1)^2}\right) \\
& =(k+1)^2\left(1+\frac{2 k+3}{(k+1)^2}\right)=(k+1)^2+2 k+3 \\
& =k^2+2 k+1+2 k+3=k^2+4 k+4=(k+2)^2 \\
& =(k+1+1)^2
\end{aligned}
$
$\therefore \mathrm{P}(\mathrm{k}+1)$ is also true whenever $\mathrm{P}(\mathrm{k})$ is true
Hence, by the principle of mathematical induction, $P(n)$ is also true for all $n \in N$.

Question 6.
$n^3-n$ is divisible by 6 , for each natural number $n \geq 2$.
Solution:
Let $P(n): n^3-n$
Step 1 :
$P(2): 2^3-2=6$ which is divisible by 6 . So it is true for $P(2)$.
Step 2 :
$P(A): \mathrm{k}^3-\mathrm{k}=6 \lambda$. Let it is be true for $\mathrm{k} \geq 2$ $\Rightarrow \mathrm{k}^3=6 \lambda+\mathrm{k}$
Step 3 :
$
\begin{aligned}
& \mathrm{P}(\mathrm{k}+1)=(\mathrm{k}+1)^3-(\mathrm{k}+1) \\
& =\mathrm{k}^3+1+3 \mathrm{k}^2+3 \mathrm{k}-\mathrm{k}-1=\mathrm{k}^3-\mathrm{k}+3\left(\mathrm{k}^2+\mathrm{k}\right) \\
& =\mathrm{k}^3-\mathrm{k}+3\left(\mathrm{k}^2+\mathrm{k}\right)=6 \lambda+\mathrm{k}-\mathrm{k}+3\left(\mathrm{k}^2+\mathrm{k}\right) \\
& =6 \lambda+3\left(\mathrm{k}^2+\mathrm{k}\right)[\text { from (i) ] }
\end{aligned}
$
We know that $3\left(k^2+k\right)$ is divisible by 6 for every value of $k \in N$.
Hence $P(k+1)$ is true whenever $P(k)$ is true.
Question 7.
For any natural number $n, 7^{\mathrm{n}}-2^{\mathrm{n}}$ is divisible by 5 .
Solution:
Let $P(n): 7^n-2^n$
Step 1 :
$P(1): 7^1-2^1=5 \lambda$ which is divisible by 5 . So it is true for $P(1)$.
Step 2 :
$P(\mathrm{k}): 7^{\mathrm{k}}-2^{\mathrm{k}}=5 \lambda$. Let it be true for $\mathrm{P}(\mathrm{k})$

Step 3 :
$
\mathrm{P}(\mathrm{k}+1)=7^{\mathrm{k}+1}-2^{\mathrm{k}+1}
$
$
\begin{aligned}
& =7^{k+1}+7^k \cdot 2-7^k \cdot 2-2^{k+1} \\
& =\left(7^{k+1}-7^k \cdot 2\right)+\left(7^k \cdot 2-2^{k+1}\right) \\
& =7^k(7-2)+2 \cdot\left(7^k-2^k\right) \\
& =5.7^k+2.5 \lambda \\
& =5\left(7^k+2 \lambda\right) \text { which is divisible by } 5 .
\end{aligned}
$
So, it is true for $\mathrm{P}(\mathrm{k}+1)$
Hence, $P(k+1)$ is true whenever $P(k)$ is true.

Question 8.
$n^2<2^n$, for all natural numbers $n \geq 5$.
Solution:
Let $\mathrm{P}(\mathrm{n}): \mathrm{n}^2<2^{\mathrm{n}}$ for all natural numbers, $\mathrm{n} \geq 5$
Step 1 :
$P(5): 1^5<2^5 \Rightarrow 1<32$ which is true for $P(5)$
Step 2:
$P(k): k^2<2 k$. Let it be true for $k \in N$
Step 3 :
$\mathrm{P}(\mathrm{k}+1):(\mathrm{k}+1)^2<2^{\mathrm{k}+1}$
From Step 2, we get $\mathrm{k}^2<2^{\mathrm{k}}$
$\Rightarrow \mathrm{k}^2<2^{\mathrm{k}+1}<2 \mathrm{k}+2 \mathrm{k}+1$
$\Rightarrow \quad(k+1)<2^k+2 k+2 k+1$
Since $(2 k+1)<2^k$
So $\quad k^2+2 k+1<2^k+2^k$
$\Rightarrow \quad k^2+2 k+1 \quad<2.2^k$
$\Rightarrow \quad k^2+2 k+1 \quad<2^{k+1}$
From eqn. (i) and (ii), we get $(\mathrm{k}+1)^2<2^{\mathrm{k}+1}$
Hence, $P(k+1)$ is true whenever $P(k)$ is true for $k \in N, n \geq 5$.

Question 9.
In $2 n<(n+2)$ ! for all natural number $n$.
Solution:
Let $\mathrm{P}(\mathrm{n}): 2 \mathrm{n}<(\mathrm{n}+2)$ ! for all $\mathrm{k} \in \mathrm{N}$.

Question 10.
$
1+5+9+\ldots+(4 \mathrm{n}-3)=\mathrm{n}(2 \mathrm{n}-1), \forall \mathrm{n} \in \mathrm{N} \text {. }
$
Solution:
Let $\mathrm{P}(\mathrm{n}): 1+5+9+\ldots+(4 \mathrm{n}-3)=\mathrm{n}(2 \mathrm{n}-1), \forall \mathrm{n} \in \mathrm{N}$
Step 1:
$P(1): 1=1(2.1-1)=1$ which is true for $P(1)$
Step 2:
$\mathrm{P}(\mathrm{k}): 1+5+9+\ldots+(4 \mathrm{k}-3)=\mathrm{k}(2 \mathrm{k}-1)$. Let it be true.
Step 3:
$
\begin{aligned}
& \mathrm{P}(\mathrm{k}+1): 1+5+9+\ldots+(4 \mathrm{k}-3)=\mathrm{k}(4 \mathrm{k}+1) \\
& =\mathrm{k}(2 \mathrm{k}-1)+(4 \mathrm{k}+1)=2 \mathrm{k}^2-\mathrm{k}+4 \mathrm{k}+1 \\
& =2 \mathrm{k}^2+3 \mathrm{k}+1=2 \mathrm{k}^2+2 \mathrm{k}+\mathrm{k}+1 \\
& =2 \mathrm{k}(\mathrm{k}+1)+1(\mathrm{k}+1)=(2 \mathrm{k}+1)(\mathrm{k}+1) \\
& =(\mathrm{k}+1)(2 \mathrm{k}+2-1)=(\mathrm{k}+1)[2(\mathrm{k}+1)-1]
\end{aligned}
$
Which is true for $P(k+1)$.
Hence, $P(k+1)$ is true whenever $P(k)$ is true.