Mathematical Induction (Part-1)| Introduction to Algebra | Unit-3 (S.N.Dey)

 Brief Overview of  Algebra Syllabus | Unit-3 (S.N.Dey)


Marks: 37                                  


1. Principle of Mathematical Induction:  [06 Periods]

Process of the proof by induction, motivating the application of the method by looking at natu ral numbers as the least inductive subset of real numbers. The principle of mathematical induction and simple applications.

2. Complex Numbers and Quadratic Equations: [10 Periods]

Need for complex numbers, especially V-1, to be motivated by inability to solve every quadratic equation. Brief description of algebraic properties of complex numbers. Argand plane and polar representation of complex numbers.

Statement of Fundamental Theorem of Algebra, solution of quadratic equations in the complex number system. Square root of a complex number, cube roots of unity and their properties.

3. Linear Inequalities:   [10 Periods]

Linear inequalities. Algebraic solutions of linear inequalities in one variable and their repre sentation on the number line. Graphical solution of linear inequalities in two variables. Solution of system of linear inequalities in two variables-graphically. Inequalities involving modulus function.

4. Permutations & Combinations:  [12 Periods ]

Fundamental principle of counting. Factorial n(n!). Permutations and combinations, deriva tion of formulae and their connections, simple applications.

5. Binomial Theorem:  [ 08 Periods ]

History, statement and proof of the binomial theorem for positive integral indices. Pascal's triangle, general and middle term in binomial expansion, simple applications.

6. Sequence and Series : [10 Periods]

Sequence and series. Arithmetic progression (A.P.), arithmetic mean (A.M.). Geometric progression (G.P.), general term of a G.P., sum of $\,n\,$ terms of a G.P., geometric mean (G.M.), relation between A.M. and G.M. Arithmetic/geometric series, infinite G.P. and its sum, Sum to n terms of the special series $\,\sum{n}$, and $\,\,\sum{n^2},\,\,\sum{n^3}.$


Mathematical Induction (Part-1)| Introduction to Algebra | Unit-3 (S.N.Dey)



Exercise: 3


Short Answer Type Questions : [4 marks ]

For all $\,\, n \in \mathbb N,\,\,$ prove by principle of mathematical induction that, 

1. (i) $\,1+3+5+.....+(2n-1)=n^2$

Sol. Let us suppose that the mathematical statement is given by : $\,P(n).$   
That means $\,\,P(n): 1+3+5+.....+(2n-1)=n^2$  
Now, $\,P(1): 1=1^2; \,$ so that $\,P(1)\,$ is true.  
Let $\,P(m)\,$ is true.  
$\therefore P(m)=1+3+5+...+(2m-1)=m^2  \\ ~~~~~~~~~~~~~~~~~~~\cdots(1)$   

Now, $\,\,P(m+1): \\1+3+5+...+(2m-1)+[2(m+1)-1]\\=m^2+(2m+2-1)\,\,[\text{By (1)}] \\=m^2+2m+1\\=(m+1)^2$

Now , since $\,P(1)\,$ is true and $\,\,P(m)\,$ is true $\Rightarrow P(m+1)\,$ is true.   
So, from the method of mathematical induction, we can say $\,\,P(n)\,$ is true for all $\,n.\,$

1(ii) $\,\,a+(a+d)+(a+2d)+ ............\text{to n terms}=\frac n2[2a+(n-1)d]$

Sol. Let $\,\,P(n): a+(a+d)+(a+2d)+ ............\text{to n terms}=\frac n2[2a+(n-1)d] \cdots(1)$

Clearly, $\,n-$th term of left hand side of $\,P(n)\,$ is $\,\,a+(n-1)d.$ 

So, (1) can be rewritten as : $\,\,P(n): a+(a+d)+(a+2d)+ ............+\{a+(n-1)d\}=\frac n2[2a+(n-1)d] \cdots(2)$

Now, $\,\,P(1): a=\frac 12[2a+(1-1)d]\,\,$ and so, $\,P(1)\,$ is true.     
Now, suppose that the statement $\,\,P(m)\,$ is true. 

So, $\,\,P(m): a+(a+d)+(a+2d)+ ............+\{a+(m-1)d\}=\frac m2[2a+(m-1)d]$

Now, $\,\,P(m+1): a+(a+d)+(a+2d)+ ...........+\{a+(m-1)d\}+\{a+(m+1-1)d\}\\=\frac m2[2a+(m-1)d]+\{a+(m+1-1)d\}\\=ma+\frac m2(m-1)d+a+md\\=ma+m^2d/2-md/2+a+md\\=(m+1)a+md\left(\frac{m-1}{2}+1\right)\\=(m+1)a+md \frac{m+1}{2}\\=\frac{m+1}{2}[2a+md]\\=\frac{m+1}{2}[2a+(m+1-1)d]$

Since, $\,\,P(1)\,$ is true and $\,P(m)\,$ is true $\implies P(m+1)\,$ is true.  
So, from the method of mathematical induction, we can say $\,\,P(n)\,$ is true for all $\,n.\,$

1(iii) $\,\sin x+\sin {2x}+\sin{3x}+ \cdots +\sin{nx}\\=\frac{\sin{\frac{n+1}{2}x}\sin\frac{nx}{2}}{\sin \frac x2}$

Sol. Let $\,P(n): \sin x+\sin {2x}+\sin{3x}+ \cdots \\+\sin{nx}=\frac{\sin{\frac{n+1}{2}x}\sin\frac{nx}{2}}{\sin \frac x2}  $                                                         
be a mathematical statement.

Now, $\,\,P(1): \sin x=\frac{\sin{\frac{1+1}{2}x}.{\sin \frac{1.x}{2}}}{\sin \frac x2}\,\,$ and so, $\,P(1)\,$ is true. 

Let the statement $\,\,P(m)\,$ is true. 

Hence, $\,P(m): \sin x+\sin {2x}+\sin{3x} \\ + \cdots +\sin{(mx)}=\frac{\sin{\frac{m+1}{2}x}\sin\frac{mx}{2}}{\sin \frac x2}$

Now, $\,P(m+1): \sin x+\sin {2x}+\sin{3x}\\+ \cdots +\sin{(mx)}+\sin{(m+1)x}\\=\frac{\sin{\frac{m+1}{2}x}\sin\frac{mx}{2}}{\sin \frac x2}+\sin{(m+1)x} \\=\frac{\sin{\frac{m+1}{2}x}\sin\frac{mx}{2}}{\sin \frac x2}+2 \sin {\frac{m+1}{2}}x . \cos {\frac{m+1}{2}}x\\=\sin {\frac{m+1}{2}}x \left(\frac{\sin(mx/2)+2\sin(x/2)\cos((m+1)/2)x}{\sin(x/2)}\right) \\=\sin {\frac{m+1}{2}}x \left(\frac{\sin(mx/2)+\sin(m+2)x/2)-\sin{(mx/2)}}{\sin(x/2)}\right)\\ = \sin {\frac{m+1+1}{2}}x \left(\frac{\sin(m+1)x/2}{\sin(x/2)}\right)$

Since, $\,\,P(1)\,$ is true and $\,P(m)\,$ is true $\implies P(m+1)\,$ is true.  
So, from the method of mathematical induction, we can say $\,\,P(n)\,$ is true for all $\,n.\,$

1(iv)$\,\sin x+\sin {2x}+\sin{3x}+ \cdots \sin{(2n-1)x}\\=\frac{\sin^2{nx}}{\sin x}$

Sol. $P(n):\,\,\sin x+\sin {2x}+\sin{3x}\\+ \cdots +\sin{(2n-1)x}=\frac{\sin^2{nx}}{\sin x}$

Now, $\,\,P(1): \sin x=\frac{\sin^2{(1.x)}}{\sin x}$

Let the mathematical statement $\,\,P(m)\,$ is true.

So, $P(m):\,\,\sin x+\sin {2x}+\sin{3x}\\+ \cdots +\sin{(2m-1)x}=\frac{\sin^2{mx}}{\sin x}$

Now, $P(m+1):\,\,\sin x+\sin {2x}+\sin{3x}\\+ \cdots+\sin{(2m-1)x}+ \sin\{2(m+1)-1\}x\\=\frac{\sin^2{mx}}{\sin x}+\sin\{(2m+1)x\}\\=\frac{\sin^2{mx}+\sin\{(2m+1)x\}\sin x}{\sin x}\\=\frac{\sin^2(mx)+\frac 12[\cos(2mx)-\cos 2(m+1)x]}{\sin x}\\=\frac{2\sin^2(mx)+[\cos(2mx)-\cos 2(m+1)x]}{2\sin x} \\=\frac{1-\cos{2mx}+[\cos(2mx)-\cos 2(m+1)x]}{2\sin x}\\= \frac{1-\cos 2(m+1)x}{2\sin x}\\=\frac{2 \sin^2{(m+1)x}}{2\sin x}\\=\frac{ \sin^2{(m+1)x}}{\sin x}$

1(v) $\,\,(\cos \theta+\sin \theta)^n=\cos{n\theta}+i \sin{n \theta},\,i=\sqrt{-1}$

Sol. Let $\,\,P(n):(\cos \theta+\sin \theta)^n=\cos{n\theta}+i \sin{n \theta}$

Now, $\,\,P(1): (\cos \theta+i\sin \theta)^1=\cos \theta +i\sin \theta \,\,$ and so $\,P(1)\,$ is true.

Now, suppose that the statement $\,\,P(m)\,$ is true. 

So, $\,\,P(m):(\cos \theta+\sin \theta)^m=\cos{m\theta}+i \sin{m \theta} \cdots(1)$

Now, $\,\,P(m+1):(\cos \theta+\sin \theta)^{m+1}\\=(\cos \theta+\sin \theta)^{m}(\cos \theta+\sin \theta)\\=(\cos{m\theta}+i \sin{m \theta})(\cos \theta +i \sin \theta)\,\,[\text{By (1)}]\\=(\cos{m\theta}\cos\theta-\sin{m \theta}\sin \theta)\\+i(\sin m\theta\cos \theta+\cos m\theta\sin \theta)\\=\cos(m\theta+\theta)+i\sin(m\theta+\theta) \\=\cos{(m+1)\theta}+i \sin{(m+1) \theta}$

Since, $\,\,P(1)\,$ is true and $\,P(m)\,$ is true $\implies P(m+1)\,$ is true.  
So, from the method of mathematical induction, we can say $\,\,P(n)\,$ is true for all $\,n.\,$


1(vi) $\,\frac 12+\frac 14+\frac 18+\cdots+\frac {1}{2^n}=1-\frac{1}{2^n}$

Sol. Let $\,\,P(n): \frac 12+\frac 14+\frac 18+\cdots+\frac {1}{2^n}=1-\frac{1}{2^n}$

Now, $\,\,P(1): \frac 12=1-\frac{1}{2^1},\,$ and so $\,P(1)\,$ is true.

Next, suppose that the statement $\,\,P(m)\,$ is true. 

So, $\,\,P(m): \frac 12+\frac 14+\frac 18+\cdots+\frac {1}{2^m}=1-\frac{1}{2^m}$

Now, $\,\,P(m+1)\\=\frac 12+\frac 14+\frac 18+\cdots+\frac {1}{2^m}+\frac{1}{2^{m+1}}\\=1-\frac{1}{2^m}+\frac{1}{2^{m+1}}\\=1-\frac1{2^m} \left(1-\frac12\right)\\=1-\frac{1}{2^m}.\frac12\\=1-\frac{1}{2^{m+1}}$

Since, $\,\,P(1)\,$ is true and $\,P(m)\,$ is true $\implies P(m+1)\,$ is true.  
So, from the method of mathematical induction, we can say $\,\,P(n)\,$ is true for all $\,n.\,$

1(vii) $\,a+ar+ar^2+.....\text{to n terms}=a.\frac{r^n-1}{r-1}\,\,[r \neq 1]$

Sol. Let $\,\,P(n): a+ar+ar^2+.....\text{to n terms}\\=a.\frac{r^n-1}{r-1},\,\,[r \neq 1]$

Now, $\,\,P(1): a=a.\frac{r^1-1}{r-1},\,[r \neq 1]$ and so $\,P(1)\,$ is true.

Clearly, the n-th term of left hand side is: $\,ar^{n-1}.$

Next, suppose that the statement $\,\,P(m)\,$ is true. 

So, $\,\,P(m): a+ar+ar^2+.....+ar^{m-1}\\=a.\frac{r^m-1}{r-1}\,\,[r \neq 1] \cdots(1)$

Now, $\,\,P(m+1): a+ar+ar^2+.....+ar^{m-1}+ar^{m+1-1}\\=a.\frac{r^m-1}{r-1}+ar^{m}\,\,[\text{By (1)}]\\=\frac{ar^m-a+ar^{m+1}-ar^m}{r-1}\\=\frac{ar^{m+1}-a}{r-1}\\=a.\frac{r^{m+1}-1}{r-1}$

Since, $\,\,P(1)\,$ is true and $\,P(m)\,$ is true $\implies P(m+1)\,$ is true.  
So, from the method of mathematical induction, we can say $\,\,P(n)\,$ is true for all $\,n.\,$

1(viii) $\,1^2+3^2+5^2+\cdots+(2n-1)^2=\frac n3(4n^2-1)$

Sol. Let $\,\,P(n): 1^2+3^2+5^2\\+\cdots+(2n-1)^2=\frac n3(4n^2-1)$

Now, $\,\,P(1): 1^2=\frac 13(4 \times 1^2-1)$ and so $\,P(1)\,$ is true.

Next, suppose that the statement $\,\,P(m)\,$ is true. 

So, $\,\,P(m): 1^2+3^2+5^2\\+\cdots+(2m-1)^2=\frac m3(4m^2-1) \cdots(1)$

Now, $\,\,P(m+1): 1^2+3^2+5^2\\+\cdots+(2m-1)^2+[2(m+1)-1]^2\\=\frac m3(4m^2-1)+[2m+1]^2\,\,[\text{By (1)}]\\=\frac m3(2m+1)(2m-1)+(2m+1)^2\\=(2m+1)\left[\frac{m(2m-1)}{3}+2m+1\right]\\=(2m+1)\frac{2m^2-m+6m+3}{3}\\=(2m+1)\frac{2m^2+5m+3}{3}\\=(2m+1)\frac{(2m+3)(m+1)}{3}\\=\frac{(m+1)(4m^2+8m+3)}{3}\\=\frac{(m+1)[4(m+1)^2-1]}{3}$

Since, $\,\,P(1)\,$ is true and $\,P(m)\,$ is true $\implies P(m+1)\,$ is true.  
So, from the method of mathematical induction, we can say $\,\,P(n)\,$ is true for all $\,n.\,$

1(ix) $\,\,1.2+2.3+3.4+\cdots+n(n+1)\\=\frac 13n(n+1)(n+2)$

Sol. Let $\,\,P(n): 1.2+2.3+3.4+\cdots+n(n+1)\\=\frac 13n(n+1)(n+2) \cdots(1)$

Now, $\,\,P(1): 1.2=\frac 131(1+1)(1+2)\,\,$ and so, $\,P(1)\,$ is true.     
Now, suppose that the statement $\,\,P(m)\,$ is true. 

So, $\,\,P(m): 1.2+2.3+3.4+\cdots+m(m+1)\\=\frac 13m(m+1)(m+2)$

Now, $\,\,P(m+1):1.2+2.3+3.4+\cdots+m(m+1)+(m+1)(m+1+1)\\=\frac 13m(m+1)(m+2)+(m+1)(m+2)\,\,[\text{By (1)}]\\=(m+1)(m+2)(\frac m3+1)\\=\frac 13(m+1)(m+1+1)(m+1+2)$

Since, $\,\,P(1)\,$ is true and $\,P(m)\,$ is true $\implies P(m+1)\,$ is true.  
So, from the method of mathematical induction, we can say $\,\,P(n)\,$ is true for all $\,n.\,$


1(x) $\,\, 2^2+5^2+8^2+\cdots \text{to n terms}\\=\frac n2(6n^2+3n-1)$

Sol. Let $\,\,P(n): 2^2+5^2+8^2+\cdots \text{to n terms}\\=\frac n2(6n^2+3n-1) $

Clearly, the n-th term of left hand side is: $\,\{2+(n-1)\times 3\}^2=(3n-1)^2$

Now, $\,\,P(1): 2^2=\frac 12(6.1^2+3.1-1)\,\,$ and so, $\,P(1)\,$ is true.     
Now, suppose that the statement $\,\,P(m)\,$ is true. 

So, $\,\,P(m): 2^2+5^2+8^2+\cdots +(3m-1)^2\\=\frac m2(6m^2+3m-1) \cdots(1)$

Now, $\,\,P(m+1): 2^2+5^2+8^2+\cdots \\+(3m-1)^2+(3 \overline{m+1}-1)^2\\=\frac m2(6m^2+3m-1)+(3m+3-1)^2\,\,[\text{By (1)}]\\=\frac m2(6m^2+3m-1)+(3m+2)^2\\=\frac 12[6m^3+3m^2-m+18m^2+24m+8]\\=\frac 12[6m^3+21m^2+23m+8]$

Moreover, $\,\,\frac{(m+1)}{2}[6(m+1)^2+3(m+1)-1]\\=\frac{m+1}{2}[6m^2+15m+8]\\=\frac 12[6m^3+21m^2+23m+8]$

Therefore, $\,\,2^2+5^2+8^2+...\\+(3m-1)^2+[3(m+1)-1]^2\\=\frac{m+1}{2}\{6(m+1)^2+3(m+1)-1\}$

Since, $\,\,P(1)\,$ is true and $\,P(m)\,$ is true $\implies P(m+1)\,$ is true.  
So, from the method of mathematical induction, we can say $\,\,P(n)\,$ is true for all $\,n.\,$

1(xi) $\,\, \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\cdots+\frac{1}{n(n+1)}=\frac{n}{n+1}$

Sol. Let $\,\,P(n): \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\cdots+\frac{1}{n(n+1)}=\frac{n}{n+1}$

Now, $\,\,P(1):\frac{1}{1.2}=\frac{1}{1+1}\,\,$ and so, $\,P(1)\,$ is true.     
Now, suppose that the statement $\,\,P(m)\,$ is true. 

So, $\,\,P(m): \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\cdots+\frac{1}{m(m+1)}=\frac{m}{m+1} \cdots(1)$

Now, $\,\,P(m+1): \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\cdots\\+\frac{1}{m(m+1)}+\frac{1}{(m+1)(m+2)}\\=\frac{m}{m+1}+\frac{1}{(m+1)(m+2)} \,\,[\text{By (1)}]\\=\frac{1}{m+1}[m+\frac{1}{m+2}]\\=\frac{m^2+2m+1}{(m+1)(m+2)}\\=\frac{(m+1)^2}{(m+1)(m+2)}\\=\frac{m+1}{m+2}\\=\frac{m+1}{m+1+1}$

Since, $\,\,P(1)\,$ is true and $\,P(m)\,$ is true $\implies P(m+1)\,$ is true.  
So, from the method of mathematical induction, we can say $\,\,P(n)\,$ is true for all $\,n.\,$

1(xii) $\,\, \frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\cdots \text{to n terms}\\=\frac{n}{3n+1}$

Sol. Let $\,\,P(n): \frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}\\+\cdots \text{to n terms}=\frac{n}{3n+1}$

Clearly, the n-th term of left hand side is:   

$\,\, \frac{1}{(1+(n-1)\times 3)(4+(n-1) \times 3)}=\frac{1}{(3n-2)(3n+1)}$

Now, $\,\,P(1):\frac{1}{1.4}=\frac{1}{3.1+1}\,\,$ and so, $\,P(1)\,$ is true.     
Now, suppose that the statement $\,\,P(m)\,$ is true. 

So, $\,\,P(m):  \frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\cdots+ \frac{1}{(3m-2)(3m+1)}=\frac{m}{3m+1}\cdots(1)$

Now, $\,\,P(m+1): \frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\cdots\\+ \frac{1}{(3m-2)(3m+1)}+\frac{1}{(3\overline{m+1}-2)(3\overline{m+1}+1)}\\=\frac{m}{3m+1}+\frac{1}{(3\overline{m+1}-2)(3\overline{m+1}+1)}\,\,\,[\text{By (1)}]\\=\frac{1}{3m+1}+\frac{1}{(3m+1)(3m+4)}\\=\frac{m(3m+4)+1}{(3m+1)(3m+4)}\\=\frac{(3m+1)(m+1)}{(3m+1)(3m+4)}\\=\frac{m+1}{3(m+1)+1}$

Since, $\,\,P(1)\,$ is true and $\,P(m)\,$ is true $\implies P(m+1)\,$ is true.  
So, from the method of mathematical induction, we can say $\,\,P(n)\,$ is true for all $\,n.\,$

1(xiii) $\,\, (1-\frac12)(1-\frac13)(1-\frac14)....(1-\frac{1}{n+1})\\=\frac{1}{n+1}$

Sol. Let $\,\,P(n): (1-\frac12)(1-\frac13)(1-\frac14)....(1-\frac{1}{n+1})\\=\frac{1}{n+1}$
 

Now, $\,\,P(1):1-\frac 12=\frac 12=\frac{1}{1+1}\,\,$ and so, $\,P(1)\,$ is true.     
Now, suppose that the statement $\,\,P(m)\,$ is true. 

So, $\,\,P(m):  (1-\frac12)(1-\frac13)(1-\frac14)....(1-\frac{1}{m+1})=\frac{1}{m+1}\cdots(1)$

Now, $\,\,P(m+1): (1-\frac12)(1-\frac13)(1-\frac14) \cdots \\ \cdots(1-\frac{1}{m+1})(1-\frac{1}{m+1+1})\\=\frac{1}{m+1}.\frac{m+1}{m+2} \,\,\,[\text{By (1)}]\\=\frac{1}{m+2}\\=\frac{1}{m+1+1}$

Since, $\,\,P(1)\,$ is true and $\,P(m)\,$ is true $\implies P(m+1)\,$ is true.  
So, from the method of mathematical induction, we can say $\,\,P(n)\,$ is true for all $\,n.\,$

1(xiv) $\,\, 1+\frac{1}{1+2}+\frac{1}{1+2+3}+\cdots+\frac{1}{1+2+3+\cdots+n}\\=\frac{2n}{n+1}$

Sol. Let $\,\,P(n): 1+\frac{1}{1+2}+\frac{1}{1+2+3}+\cdots\\+\frac{1}{1+2+3+\cdots+n}=\frac{2n}{n+1}$
 

Now, $\,\,P(1):1=\frac{2 \times 1}{1+1}\,\,$ and so, $\,P(1)\,$ is true.     
Now, suppose that the statement $\,\,P(m)\,$ is true. 

So, $\,\,P(m): 1+\frac{1}{1+2}+\frac{1}{1+2+3}+\cdots\\+\frac{1}{1+2+3+\cdots+m}=\frac{2m}{m+1}\cdots(1)$

Now, $\,\,P(m+1):  1+\frac{1}{1+2}+\frac{1}{1+2+3}+\cdots\\+\frac{1}{1+2+3+\cdots+m}+\frac{1}{1+2+3+\cdots+m+m+1}\,\,\,\\=\frac{2m}{m+1}+\frac{1}{1+2+3+\cdots+m+m+1}\,\,[\text{By (1)}]\\=\frac{2m}{m+1}+\frac{1}{\frac{m+1}{2}(m+1+1)}\\=\frac{2m}{m+1}+\frac{2}{(m+1)(m+2)}\\=\frac{2}{m+1}[m+\frac{1}{m+2}]\\=\frac{2}{m+1} \times \frac{m^2+2m+1}{m+2}\\=\frac{2}{m+1} \times \frac{(m+1)^2}{m+2}\\=\frac{2(m+1)}{m+1+1}$

Since, $\,\,P(1)\,$ is true and $\,P(m)\,$ is true $\implies P(m+1)\,$ is true.  
So, from the method of mathematical induction, we can say $\,\,P(n)\,$ is true for all $\,n.\,$

1(xv) $\,\, 1.1!+2.2!+3.3!+\cdots+n.n!\\=(n+1)!-1$

Sol. Let $\,\,P(n):  1.1!+2.2!+3.3!+\cdots+n.n!\\=(n+1)!-1$
 
Now, $\,\,P(1):1.1!=(1+1)!-1\,\,$ and so, $\,P(1)\,$ is true.     
Now, suppose that the statement $\,\,P(m)\,$ is true. 

So, $\,\,P(m): 1.1!+2.2!+3.3!+\cdots+m.m!\\=(m+1)!-1\cdots(1)$

Now, $\,\,P(m+1):  1.1!+2.2!+3.3!+\cdots\\+m.m!+(m+1).(m+1)!\\=(m+1)!-1+(m+1).(m+1)!\,\,[\text{By (1)}]\\=(m+1)!(1+m+1)-1\\=(m+2)(m+1)!-1\\=(m+2)!-1\\=(m+1+1)!-1$

Since, $\,\,P(1)\,$ is true and $\,P(m)\,$ is true $\implies P(m+1)\,$ is true.  
So, from the method of mathematical induction, we can say $\,\,P(n)\,$ is true for all $\,n.\,$


1(xvi) $\,\, 2^{3n}-1 \,\,\text{is divisible by}\,\,7.$

Sol. Let $\,\,P(n):  2^{3n}-1 \,\,\text{is divisible by}\,\,7.$
 

Now, $\,\,P(1):2^{3.1}-1=8-1=7\,\,$ and so, $\,P(1)\,$ is true.     
Now, suppose that the statement $\,\,P(m)\,$ is true. 

So, $\,\,P(m): 2^{3m}-1 \,\,\text{is divisible by}\,\,7 \\ \therefore2^{3m}-1=7k,\,\,\text{where,}\,\,k \in \mathbb Z\cdots(1)$

Now, $\,\,P(m+1):  2^{3(m+1)}-1\\=2^{3m+3}-1\\=8 \times 2^{3m}-1\\=8(7k+1)-1,\,\,\text{By (1)}\\=56k+7\\=7(8k+1)$ 

And so, $\,\,2^{3(m+1)}-1\,\,$ is divisible by $\,7.\,\,\,[\text{Since,}\,\,k \in \mathbb Z, 8k+7 \in \mathbb Z]$ 

Since, $\,\,P(1)\,$ is true and $\,P(m)\,$ is true $\implies P(m+1)\,$ is true.  
So, from the method of mathematical induction, we can say $\,\,P(n)\,$ is true for all $\,n.\,$

1(xvii) $\,\, 4^n+15n-1 \,\,\text{is a multiple of}\,\,9.$

Sol. Let $\,\,P(n): 4^n+15n-1 \,\,\text{is a multiple of}\,\,9.$
 

Now, $\,\,P(1):4^1+15 \times 1-1=18 \\ \text{which is a multiple of}\,\,9\,\,\,$           
 and so, $\,P(1)\,$ is true.     
Now, suppose that the statement $\,\,P(m)\,$ is true. 

So, $\,\,P(m): 4^m+15m-1 \,\,\text{is a multiple of}\,\,9 \\ \therefore 4^m+15m-1=9k,\,\,\text{where,}\,\,k \in \mathbb Z\\ \Rightarrow 4^m=9k-15m+1 \cdots(1) $

Now, $\,\,P(m+1):  4^{m+1}+15(m+1)-1\\=4 \times 4^{m}+15m+14\\=4 \times (9k-15m+1)+15m+14\,\,[\text{By (1)}]\\=36k-60m+4+15m+14\\=36k-45m+18\\=9(4k-5m+2)$ 

And so, $\,\,4^{m+1}+15(m+1)-1\,\,$ is a multiple of $\,9.\,\,\,[\text{Since,}\,\,k \in \mathbb Z, m \in \mathbb N,\,4k-5m+2\in \mathbb Z]$ 

Since, $\,\,P(1)\,$ is true and $\,P(m)\,$ is true $\implies P(m+1)\,$ is true.  
So, from the method of mathematical induction, we can say $\,\,P(n)\,$ is true for all $\,n.\,$


1(xviii) $\,\, 2.7^n+3.5^n-5 \,\,\text{is divisible by}\,\,24.$

Sol. Let $\,\,P(n): 2.7^n+3.5^n-5 \,\,\text{is divisible by}\,\,24.$
 

Now, $\,\,P(1): 2.7^1+3.5^1-5=24 \,\,\text{is divisible by}\,\,24\,\,\,$ and so, $\,P(1)\,$ is true.     
Now, suppose that the statement $\,\,P(m)\,$ is true. 

So, $\,\,P(m):  2.7^m+3.5^m-5 \,\,\text{is divisible by}\,\,24 \\ \therefore 2.7^m+3.5^m-5=24k,\,\,\text{where,}\,\,k \in \mathbb Z \cdots(1) $

Now, $\,\,P(m+1):  2.7^{m+1}+3.5^{m+1}-5\\=14.7^m+15.5^m-5\\=24k+12(7^m+5^m)\,\,[\text{By (1)}]\\=24k+24k_1 \\ [\text{Since}\,\, 7^m,5^m\,\,\text{is odd,}\,\,7^m+5^m\,\,\text{is even}=2k_1]\\=24(k+k_1)$ 

Hence, $\,2.7^{m+1}+3.5^{m+1}-5 \,\,\text{is divisible by}\,\,24.$

Since, $\,\,P(1)\,$ is true and $\,P(m)\,$ is true $\implies P(m+1)\,$ is true.  
So, from the method of mathematical induction, we can say $\,\,P(n)\,$ is true for all $\,n.\,$

1(xix) $\,\,15^{2n-1}+1\,\,$ is divisible by $\,\,16.$

Sol. Let $\,\,P(n):15^{2n-1}+1\,\,$ is divisible by $\,\,16.$

So, $\,\,P(1):15^{2\times 1-1}+1=16\,\,$ which is divisible by $\,\,16.$

Suppose, $\,\,P(m)\,$ is true. 

So, $\,\,P(m):15^{2m-1}+1\,\,$ is divisible by $\,\,16.$

Therefore, $\,\,15^{2m-1}+1=16k,\,\,k \in \mathbb Z \\ \Rightarrow15^{2m-1}=16k-1 \cdots(1)$

Now, $\,\,P(m+1): 15^{2(m+1)-1}+1\\=15^{2m+2-1}+1 \\ =15^2.15^{2m-1}+1\\=225(16k-1)+1\,\,[\text{By (1)}]\\=225 \times 16k-224\\=16(225k-14)$

So, $\,15^{2m-1}+1\,\,$ is always divisible by $\,\,16.$

Since, $\,\,P(1)\,$ is true and $\,P(m)\,$ is true $\implies P(m+1)\,$ is true.  
So, from the method of mathematical induction, we can say $\,\,P(n)\,$ is true for all $\,n.\,$


1(xx) $\,\,12^n+25^{n-1}\,\,$ is divisible by $\,\,13.$

Sol. Let $\,\,P(n):12^n+25^{n-1}\,\,$ is divisible by $\,\,13.$

So, $\,\,P(1):12^1+25^{1-1}=13\,\,$ which is divisible by $\,\,13.$

Suppose, $\,\,P(m)\,$ is true. 

So, $\,\,P(m):12^m+25^{m-1}\,\,$ is divisible by $\,\,13.$

Therefore, $\,\,12^m+25^{m-1}=13k,\,\,k \in \mathbb Z  \cdots(1)$

Now, $\,\,P(m+1): 12^{m+1}+25^{m+1-1}\\=12\times 12^m+25 \times 25^{m-1}\\=12\times 12^m+12 \times 25^{m-1}+13 \times 25^{m-1}\\=12(12^m+25^{m-1})+13 \times 25^{m-1}\\=12 \times 13k+13 \times 25^{m-1}\,\,\,[\text{By (1)}]\\=13(12k+25^{m-1})$

So, $\,12^m+25^{m-1}\,\,$ is always divisible by $\,\,13.$

Since, $\,\,P(1)\,$ is true and $\,P(m)\,$ is true $\implies P(m+1)\,$ is true.  
So, from the method of mathematical induction, we can say $\,\,P(n)\,$ is true for all $\,n.\,$

2. For what natural numbers $\,n\,$ the inequality $\,\,2^n >2n+1\,\,$ is valid?

Sol. Clearly, for $\,\,n=1,\,2^1=2> 2 \times 1+1=3\,$ which means the aforementioned inequality is not valid.  
Similarly, for $\,\,n=2,\,2^2=4>2\times 2+1 \Rightarrow 4>5\,$which means the aforesaid inequality is not valid. 

Now, for $\,\,n=3,\, 2^3>2 \times 3+1 \Rightarrow 8>7\,$ which is true.

Let $\,\,P(n): 2^n>2n+1\,\,\forall n \geq 3 \wedge n \in \mathbb N,\,\,$ and which means $\,P(3)\,$ is true. 

Now, let $\,\,P(m)\,$ is true. 

Therefore, $\,\,2^m>2m+1\cdots(1)$

Now, we have to prove , $\,\,P(m+1)\,$ is true, i.e., $\,\,2^{m+1}>2m+3.$

Now, $\,2^{m+1}=2.2^m>2(2m+1)\,\,[\text{By (1)}]\\=2m+2m+2\\>2m+3\,\,[\text{Since,}\,m \geq 3 \Rightarrow 2m+2 \geq 6+2>3]\\ \Rightarrow 2^{m+1}>2(m+1)+1$

Since, $\,\,P(3)\,$ is true and $\,P(m)\,$ is true $\implies P(m+1)\,$ is true.  
So, from the method of mathematical induction, we can say $\,\,P(n)\,$ is true for all $\,n \geq3,\,\,n \in \mathbb N.$


To continue with Long Answer Type Questions  and for  their solutions of S.N.Dey Math (Exercise-3), please click here.


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