Differentiation (Part-38) | S N De

 In the previous chapter, we have discussed Short Answer Type Questions  of Mathematical Induction of S.N.Dey. In this article, we will discuss Long Answer Type Questions [EX-3] and their solutions. So, let's start.

1 If $\,\,n \in \mathbb N,\,\,$ then by principle of mathematical in prove that, 

(i) $\,\,5^{2n+2}-24n-25\,\,$ is divisible by $\,\,576.$

Sol. Let $\,\,P(n):5^{2n+2}-24n-25\,\,$ is divisible by $\,\,576\,\,$ be a mathematical statement.

Now, $\,\,P(1): 5^{2 \times 1+2}-24 \times 1-25\\=5^4-24-25\\=625-49\\=576,\,\,\text{which is divisible by 576.}$

Hence, $\,P(1)\,$ is true. 

Let $\,P(m)\,$ be true.  

Then, $\,\,P(m):5^{2m+2}-24m-25\,\,$ is divisible by $\,\,576$  

Hence,$ \,\,5^{2m+2}-24m-25=576k,\,\,k \in \mathbb N \\ \Rightarrow 5^{2m+2}=576k+24m+25 \cdots(1)$

Now, $\,\,P(m+1): 5^{2(m+1)+2}-24(m+1)-25\\=5^{2m+2+2}-24m-24-25\\=25 \times 5^{2m+2}-24m-49\\=25(576k+24m+25)-24m-49\,\,[\text{By (1)}]\\=25 \times 576k+25 \times 24m+625-24m-49\\=25 \times 576k+24 \times 24m+24m+576\\=576(25k+m+1).$

And so, $\,5^{2(m+1)+2}-24(m+1)-25\,$ is always divisible by $\,576.$

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 $\,n \in \mathbb N.$

1 If $\,\,n \in \mathbb N,\,\,$ then by principle of mathematical in prove that, 

(ii) $\,\,10^n+3.4^{n+2}+5\,\,[n \geq 0]$ is divisible by $\,\,9.$

Sol. Let $\,\,P(n):10^n+3.4^{n+2}+5\,\,[n \geq 0]$ is divisible by $\,\,9\,\,-$ be a mathematical statement.

Now, $\,\,P(1): 10^1+3.4^{1+2}+5 \\=10+3.4^3+5\\=10+192+5\\=307,\,\,\text{which is divisible by 9.}$

Hence, $\,P(1)\,$ is true. 

Let $\,P(m)\,$ be true.  

Then, $\,\,P(m):10^m+3.4^{m+2}+5\,\,[m \geq 0]$ is divisible by $\,\,9.$

Hence,$ \,\,10^m+3.4^{m+2}+5=9k, \,\,k \in \mathbb N \\ \Rightarrow 10^m=9k-3.4^{m+2}-5 \cdots(1)$

Now, $\,\,P(m+1): 10^{m+1}+3.4^{\overline{m+1}+2}+5\\=10.10^m+3.4^{m+3}+5\\=10(9k-3.4^{m+2}-5)+12.4^{m+2}+5\\=90k-30.4^{m+2}-50+12.4^{m+2}+5\\=90k-18.4^{m+2}-45\\=9(10k-2.4^{m+2}-5)$

And so, $\,10^{m+1}+3.4^{\overline{m+1}+2}+5\,$ is always divisible by $\,9.$

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 $\,n \in \mathbb N.$

1 If $\,\,n \in \mathbb N,\,\,$ then by principle of mathematical in prove that, 

(iii) $\,\,3^{4n+1}+2^{2n+2}\,\,[n \geq 0]$ is multiple of $\,\,7.$

Sol. Let $\,\,P(n):\,3^{4n+1}+2^{2n+2}\,\,[n \geq 0]$ is multiple of $\,\,7-$ be a mathematical statement.

Now, $\,\,P(1): 3^{4\times 1+1}+2^{2 \times 1 +2}\\=3^5+2^4\\=243+16=259\,\,\text{which is a multiple of }\,\,7.$

Hence, $\,P(1)\,$ is true. 

Let $\,P(m)\,$ be true.  

Then, $\,\,P(m):\,3^{4m+1}+2^{2m+2}\,\,[m \geq 0]$ is multiple of $\,\,7.$

Hence,$ \,\,3^{4m+1}+2^{2m+2} =7k,\,\,\,\,k \in \mathbb N \\ \Rightarrow 3^{4m+1}=7k-2^{2m+2} \cdots(1)$

Now, $\,\,P(m+1): 3^{4(m+1)+1}+2^{2(m+1)+2}\\=3^{4m+1+4}+2^{2m+4}\\=3^{4m+1}.3^4+2^2.2^{2m+2}\\=81.3^{4m+1}+4.2^{2m+2}\\=81.(7k-2^{2m+2})+4.2^{2m+2}\\=81 \times 7k -81.2^{2m+2}+4.2^{2m+2}\\=81 \times 7k -77.2^{2m+2}\\=7[81k-11.2^{2m+2}]$

And so, $\,3^{4(m+1)+1}+2^{2(m+1)+2}\,$ is a multiple of $\,7.$

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 $\,n \in \mathbb N.$

1 If $\,\,n \in \mathbb N,\,\,$ then by principle of mathematical in prove that, 

(iv) $\,\,3^{2n+2}-8n-9\,\,$ is divisible by $\,\,64.$

Sol. Let $\,\,P(n):3^{2n+2}-8n-9\,\,$ is divisible by $\,\,64-$be a mathematical statement.

Now, $\,\,P(1): 3^{2\times 1+2}-8 \times 1-9\\=3^{4}-17\\=64\,\,\text{which is divisible by}\,\,64.$

Hence, $\,P(1)\,$ is true. 

Let $\,P(m)\,$ be true.  

Then, $\,\,P(m):3^{2m+2}-8m-9\,\,$ is divisible by $\,\,64.$

Hence,$ \,\,3^{2m+2}-8m-9 =64k,\,\,\,\,k \in \mathbb N \\ \Rightarrow 3^{2m+2}=64k+8m+9 \cdots(1)$

Now, $\,\,P(m+1): 3^{2(m+1)+2}-8(m+1)-9\\=3^2(64k+8m+9)-8m-17 \,\,[\text{By (1)}]\\=9\times 64k+72m+81-8m-17\\=9 \times 64k+64m+64\\=64(9k+m+1)$

And so, $\,3^{2(m+1)+2}-8(m+1)-9\,$ is divisible by $\,64.$

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 $\,n \in \mathbb N.$

1 If $\,\,n \in \mathbb N,\,\,$ then by principle of mathematical in prove that,   

(v) $\,\,7+77+777+....\text{to n terms}\\=\frac{7}{81}(10^{n+1}-9n-10).$

Sol. Let $\,\,P(n):7+77+777+....\text{to n terms}\\=\frac{7}{81}(10^{n+1}-9n-10)-$

 be a mathematical statement.

Now, $\,\,P(1): 7=\frac{7}{81}(10^{1+1}-9\times 1-10)\\=\frac{7}{81}\times(100-9-10).$

Hence, $\,P(1)\,$ is true. 

Let $\,P(m)\,$ be true.  

Then, $\,\,P(m):7+77+777+....\text{to m terms}\\=\frac{7}{81}(10^{m+1}-9m-10) \\ \Rightarrow P(m): \frac 79(9+99+999+\cdots \text{ to m terms})\\=\frac{7}{81}(10^{m+1}-9m-10)\\ \Rightarrow P(m): \frac 79 [(10-1)+(10^2-1)+(10^3-1)+\cdots +(10^m-1)]\\=\frac{7}{81}(10^{n+1}-9n-10)  \cdots(1)$  

So, $\,\,P(m+1): \frac 79 [(10-1)+(10^2-1)\\+(10^3-1)+\cdots  +(10^m-1)+(10^{m+1}-1)]\\=\frac{7}{81}(10^{m+1}-9m-10)+\frac 79[10^{m+1}-1] \\ [\text{From (1)}]\\=\frac{7}{81}[(10^{m+1}-9m-10)+ 9(10^{m+1}-1)] \\=\frac{7}{81}[10^{m+1}-9m-10+9.10^{m+1}-9]\\=\frac{7}{81}[10.10^{m+1}-9m-10-9]\\=\frac{7}{81}[10^{m+1+1}-9(m+1)-10]$  

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 $\,n \in \mathbb N.$

1 If $\,\,n \in \mathbb N,\,\,$ then by principle of mathematical in prove that,   

(vi) $\,(a)\, n <2^n\,\,(b)\,3^n >n^3\,[n \geq 4] \\ (c)\,2^n<n!\,\,[n \geq 4]\,\,(d)\,3^n>2^n$

Sol. (vi-a) Let $\,\,P(n):n <2^n-$

 be a mathematical statement.

Now, $\,\,P(1): 1 <2^1 \Rightarrow 1<2$

Hence, $\,P(1)\,$ is true. 

Let $\,P(m)\,$ be true.  

Then, $\,\,P(m):m <2^m \cdots(1)$  

So, $\,\,P(m+1): m+1<2^m+1\,\,[\text{By (1)}]\\ \Rightarrow m+1< 2^{m+1} \\ [\text{Since,}\,\,2^m+1<2^{m+1},\,\,\forall m \in \mathbb N]$  

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 $\,n \in \mathbb N.$

Sol. (vi-b) Let $\,\,P(n):3^m>m^3-$

 be a mathematical statement.

Now, $\,\,P(4): 3^4=81 <4^3=64.$

Hence, $\,P(4)\,$ is true. 

Let $\,P(m)\,$ be true.  

Then, $\,\,P(m):3^m>m^3 \,\,[m \geq 4]\cdots(1)$  

So, $\,\,P(m+1): 3^{m+1}=3^m.3>3m^3\,\,[\text{By (1)}]  \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\cdots(2)$

$\,\,\text{For}\,\,\,n\geq 4,\,\,(m+1)^3\\=m^3+3m^2+3m+1 \\<m^3+m^3+m^3=3m^3 \cdots(3)$

Hence, from (2) and (3), we have,        

 $\,\,P(m+1): 3^{m+1}>3m^3>(m+1)^3, \\ ~~~~~~~~~~~~~~~~~~~~~~~~~\text{for}\,\,n \geq 4.$  

Now , since $\,P(4)\,$ 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 $\,n \in \mathbb N.$

1(vi-c)  Sol. Let $\,\,P(n)\,$ be $\,\,2^n<n!.$  

Now,$\,\,P(4)\,$ is true since $\,\,2^4=16<24=4!.$

Suppose, $\,\,P(m)\,\,$ is true, i.e., $\,2^m <m!,\,\,m \geq 4.$

Them, $\,\,2^{m+1}\\=2^m.2\\ <m!. 2\\ <m!.(m+1) \\=(m+1)!,$ 

amd so $\,\,P(m+1)\,\,$ is true. 

Thus, $\,\,P(n)\,\,$ is true $\,\forall n \geq 4,\,\,n \in \mathbb N.$

1(vi-d)  Sol. Let $\,\,P(n)\,$ be $\,\,3^n >2^n.$  

Now,$\,\,P(1)\,$ is true since $\,\,3^1=3>2^1=2$

Suppose, $\,\,P(m)\,\,$ is true, i.e., $\,3^m >2^m.\,$

Then, $\,\,P(m+1):3^{m+1}\\=3.3^m \\ >2^m.3\\>2^m.2\,\, [\text{Since,}\,\,3>2] \\=2^{m+1}$ 

and so $\,\,P(m+1)\,\,$ is true. 

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 $\,n \in \mathbb N.$

1(vii) $\,\,1+2+3+\cdots n <\frac 18(2n+1)^2$

Sol. Let $\,\,P(n)\,$ be  $\,\,1+2+3+\cdots n <\frac 18(2n+1)^2$

Now,$\,\,P(1)\,$ is true since $\,\,1 <\frac 18(2 \times 1+1)^2=\frac 98$

Suppose, $\,\,P(m)\,\,$ is true, i.e., $\,\,1+2+3+\cdots m <\frac 18(2m+1)^2 \cdots (1)$

Then, $\,\,P(m+1):1+2+3+\cdots +m +(m+1) \\ < \frac 18(2m+1)^2 +(m+1)\,\,[\text{By (1)}]\\ <\frac 18[4m^2+4m+1+8m+8]\\<\frac 18[4m^2+12m+9]\\< \frac 18(2m+3)^2 \\ <\frac 18 [2(m+1)+1]^2$ 

and so $\,\,P(m+1)\,\,$ is true. 

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 $\,n \in \mathbb N.$

1(viii) $\,\,1^2+2^2+3^2+\cdots+n^2>\frac{n^3}{3}$

Sol.  Let $\,\,P(n)\,$ be $\,\,1^2+2^2+3^2+\cdots+n^2>\frac{n^3}{3}.$

Now, $\,\,P(1): 1^2=1>\frac{1^3}{3}=\frac 13\,\,$ which is true.

Suppose, $\,\,P(m)\,\,$ is true, i.e.,  $\,\,1^2+2^2+3^2+\cdots+m^2>\frac{m^3}{3} \cdots(1)$

Then, $\,\,P(m+1):1^2+2^2+3^2+\cdots\\ +m^2 +(m+1)^2 \\ > \frac{m^3}{3} +(m+1)^2\,\,[\text{By (1)}]\\ =\frac 13[m^3+3(m^2+2m+1)]\\=\frac 13[m^3+3m^2+6m+3]\\>\frac 13[m^3+3m^2+3m+1]\\=\frac 13(m+1)^3$ 

and so $\,\,P(m+1)\,\,$ is true. 

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 $\,n \in \mathbb N.$

1(viii) $\,\,\frac 15n^5+\frac 13n^3+\frac{1}{15}. 7n\,\,$ is an integer.

Sol.  Let $\,\,P(n)\,$ be $\,\,\frac 15n^5+\frac 13n^3+\frac{1}{15}. 7n\,\,$ is an integer.

Now, $\,\,P(1): \frac 15.1^5+\frac 13.1^3+\frac{1}{15}.7 \times 1=\frac 15+\frac 13+\frac{7}{15}\\=\frac{15}{15}=1\,\,$ which is an integer.

Suppose, $\,\,P(m)\,\,$ is true, i.e.,  $\,\,\frac 15m^5+\frac 13m^3+\frac{1}{15}. 7m=k(\in  I)\,\,$ is an integer.

Then, $\,\,P(m+1):\frac 15(m+1)^5+\frac 13(m+1)^3\\+\frac{1}{15}. 7(m+1) \\=\frac 15(m^5+5m^4+10m^3+10m^2+5m+1)\\+\frac 13(m^3+3m^2+3m+1)+\frac{7}{15}m +\frac{7}{15}\\=(\frac 15m^5+\frac 13m^3+\frac{1}{15}. 7m)\\+(m^4+2m^3+2m^2+m)\\+(m^2+m)+(\frac 15+\frac 13+\frac{7}{15}),$ 

and it is an integer as each element within the braces is integer and hence $\,\,P(m+1)\,\,$ is true. 

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 $\,n \in \mathbb N.$

1(x)$\,\,\frac{n^{11}}{11}+\frac{n^5}{5}+\frac{n^3}{3}+\frac{62n}{165}\,\,\,$ is an integer.

Sol.  Let $\,\,P(n)\,$ be $\,\,\frac{n^{11}}{11}+\frac{n^5}{5}+\frac{n^3}{3}+\frac{62n}{165}\,\,\,$ is an integer.

Then, $\,\,P(1):\frac{1^{11}}{11}+\frac{1^5}{5}+\frac{1^3}{3}+\frac{62 \times 1}{165}\\ =\frac{1}{11}+\frac 15+\frac 13+\frac{62}{165} \\ =\frac{15+33+55+62}{165}\\=\frac{165}{165}\\=1$

So, $\,\,P(1)\,\,$ is true.

Let $\,P(m)\,$ be true and so $\,\,P(m)\,$ be $\,\,\frac{m^{11}}{11}+\frac{m^5}{5}+\frac{m^3}{3}+\frac{62m}{165}\,\,\,$ is an integer.

Now, $\,\,P(m+1): \,\,\frac{(m+1)^{11}}{11}+\frac{(m+1)^5}{5}\\+\frac{(m+1)^3}{3}+\frac{62(m+1)}{165} \\=\frac{1}{11}(m^{11}+\cdots +1)+\frac{1}{5}(m^5+\cdots+1)\\+\frac 13(m^3+\cdots +1)+\frac{62m}{165}+\frac{62}{165}\\=\left(\frac{m^{11}}{11}+\frac{m^5}{5}+\frac{m^3}{3}+\frac{62m}{165} \right)\\+ \left(\frac{1}{11}+\frac 15+\frac 13+\frac{62}{165} \right)+I_1 \\ \,\,\text{where, other terms in the middle are integers}\,\,(I_1)\\=\text{Integer}$  

and hence $\,\,P(m+1)\,\,$ is true. 

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 $\,n \in \mathbb N.$

1(xi) $\,\,n.1+(n-1).2+(n-2).3+\cdots\\+2.(n-1)+1.n=\frac 16n(n+1)(n+2)$

Sol.   Let $\,\,P(n)\,$ be $\,\,n.1+(n-1).2+(n-2).3+\cdots \\+2.(n-1)+1.n=\frac 16n(n+1)(n+2).$

Now, $\,\,P(1): 1.1=\frac 16\times 1\times(1+1) \times(1+2)=1,$

and so, $\,P(1)\,$ is true. 

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

and so $\,\, P(m):m.1+(m-1).2+(m-2).3+\cdots\\+2.(m-1)+1.m \\=\frac 16m(m+1)(m+2)\cdots(1)$

Now, $\,\,P(m+1): (m+1).1+(m+1-1).2\\+(m+1-2).3+\cdots+2.(m+1-1)\\+1.(m+1)=(m+1).1+(m-1+1).2\\+(m-2+1).3+\cdots+1.(m+1)\\=\{m.1+(m-1).2+(m-2).3\\+\cdots+1.m\}+\{1+2+3+\cdots+(m+1)\} \\=\frac 16(m+1)(m+1+1)(m+1+2)+\frac 12 (m+1)(m+2)\,\,[\text{By (1)}]\\=\frac 16 (m+1)(m+2)(m+3)\\=\frac 16 (m+1)\{(m+1)+1\}\{(m+1)+2\}$

and hence $\,\,P(m+1)\,\,$ is true. 

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 $\,n \in \mathbb N.$

1(xii) Prove that $\,\,\, 2 +\sqrt{2+\sqrt{2+\sqrt{2+\cdots\text{n times}}}} \\<4, \,\,\forall n \geq 1$

Sol. Let $\,P(n)\,$ be $\,\,\, 2 +\sqrt{2+\sqrt{2+\sqrt{2+\cdots\text{n times}}}} \\<4, \,\,\forall n \geq 1.$

Now, $\,P(1):\, 2 <4\,\,$ and so $\,P(1)\,\,$ is true. 

Let us suppose $\,P(m)\,$ be true .

So, $\,\,\,P(m): 2 +\sqrt{2+\sqrt{2+\sqrt{2+\cdots\text{m times}}}} <4, \,\,\forall m \geq 1.$

Now , since $\,\,\, 2 +\sqrt{2+\sqrt{2+\sqrt{2+\cdots\text{m times}}}}\\ <4, \,\,\forall m \geq 1\\ \sqrt{2 +\sqrt{2+\sqrt{2+\sqrt{2+\cdots\text{(m+1) times}}}} }\\ <\sqrt{4}=2 \\ \therefore 2+\sqrt{2 +\sqrt{2+\sqrt{2+\sqrt{2+\cdots\text{(m+1) times}}}} }\\ <2+2=4$

and hence , $\,\,P(m+1)\,$ is true. 

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 $\,n \in \mathbb N.$

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