Differentiation (Part-38) | S N De

 In the  previous article, we have discussed few Long Answer Type Questions [1(i) -(xii)]. In this article , we will discuss the remaining math solution problems and their solutions. So, without further ado, let's start. 

1(xiii) Prove that $\,\,\, \left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right) \cdots \\ \left(1-\frac{1}{(n+1)^2}\right)=\frac{n+2}{2n+2}$

Sol. Let $\,P(n)\,$ be $\,\,\, \left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right) \cdots \\  \left(1-\frac{1}{(n+1)^2}\right)=\frac{n+2}{2n+2}$

Now, $\,P(1):\, 1-\frac{1}{2^2}=\frac{1+2}{2.1+2}\,\,$ and so $\,P(1)\,\,$ is true. 

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

So, $\,\,\, P(m):\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right) \cdots \\  \left(1-\frac{1}{(m+1)^2}\right)=\frac{m+2}{2m+2}$

Now , $\,\,\, P(m+1):\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right) \cdots \\  \left(1-\frac{1}{(m+1)^2}\right)\left(1-\frac{1}{(m+2)^2}\right)\\=\frac{m+2}{2m+2}\left(1-\frac{1}{(m+2)^2}\right)\\=\frac{m+2}{2m+2} \times \frac{m^2+4m+3}{(m+2)^2}\\=\frac{(m+1)(m+3)}{2(m+1)(m+2)}\\=\frac{(m+1)+2}{2(m+1)+2}$

 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(xiv) Prove that $\,\,\, 2+222+22222+\cdots\\+22...2\{(2n-1)\,\,\text{digits}\}\\=\frac{20}{891}\left(10^{2n}-1\right)-\frac{2n}{9}$

Sol. Let $\,P(n)\,$ be $\,\,\, 2+222+22222\\+\cdots+22...2\{(2n-1)\,\,\text{digits}\}\\=\frac{20}{891}\left(10^{2n}-1\right)-\frac{2n}{9}$

Now, $\,P(1):\, 2=\frac{20}{891}\left(10^{2.1}-1\right)-\frac{2.1}{9}\,\,$ and so $\,P(1)\,\,$ is true. 

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

So, $\,\,\,P(m): 2+222+22222+\cdots\\+22...2\{(2m-1)\,\,\text{digits}\}\\=\frac{20}{891}\left(10^{2m}-1\right)-\frac{2m}{9}$

Now , $\,\,\, P(m+1):2+222+22222+\cdots\\+22...2\{(2m+1)\,\,\text{digits}\}\\=\frac{20}{891}\left(10^{2m}-1\right)-\frac{2m}{9}+\frac 29.10^{2m+1}-\frac 29\\=\frac{20}{891}\left(10^{2m+2}-1\right)-\frac 29(m+1)$

 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(xv) Prove that $\,\,\, ^{n}C_0+^{n}C_1+^{n}C_2+\cdots +^{n}C_n=2^n\,\,(n \in \mathbb N)$

Sol. Let $\,P(n)\,$ be $\,\,\, ^{n}C_0+^{n}C_1+^{n}C_2+\cdots +^{n}C_n=2^n\,\,(n \in \mathbb N)$

Now, $\,P(1):\, ^{1}C_0+^{1}C_1=1+1=2^1\,\,$ and so $\,P(1)\,\,$ is true. 

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

So, $\,\,\,P(m):\,\, ^{m}C_0+^{m}C_1+^{m}C_2+\cdots \\+^{m}C_m=2^m\,\,(m \in \mathbb N)$

Now , $\,\,\, P(m+1):\,\, ^{m+1}C_0+^{m+1}C_1+^{m+1}C_2+\cdots \\+^{m+1}C_{m+1}\\=^{m+1}C_0+\sum_{r=1}^{m+1}  [^{m+1}C_r]\\=^mC_0+\sum_{r=1}^{m}\left(^mC_r+^mC_{r-1}\right)+^{m+1}C_{m+1}\\=(^mC_0+^mC_1+^mC_2 +\cdots+^mC_m) \\+(^mC_0+^mC_1+^mC_2+\cdots+^mC_m)\\=2^m+2^m\\=2^{m+1}$

 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 $\,n \in \mathbb N.$

2. If $\,x\,$ and $\,y\,$ are two real numbers, then prove by mathematical induction that $\,(x^n-y^n)\,$ is divisible by $\,(x-y)\,$ for all $\,\,n \in \mathbb N$.

Sol. Let $\,P(n)\,$ be the mathematical statement i.e., $\,(x^n-y^n)\,$ is divisible by $\,(x-y)\,$ for all $\,\,n \in \mathbb N$.

Now, $\,P(1):\, (x^1-x^1)\,\text{is divisible by }\,(x-y)$ and so $\,P(1)\,\,$ is true. 

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

So, $\,\,x^m-y^m=k(x-y) \\ \therefore x^m=k(x-y)+y^m \cdots (1)$

Now, $\,\,P(m+1): x^{m+1}-y^{m+1}\\=x^m.x-y^m.y\\=[k(x-y)+y^m].x-y^m.y \,\,[\text{By (1)}]\\=(x-y)(kx+y^m) \cdots(2)\,\,$ 

and so from (2), we can deduce that  $\,\,P(m+1)\,$ 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 $\,n \in \mathbb N.$

3. By induction method prove that, $\,(a^n + b^n)\,$ is divisible by $\,(a+b)\,$ when $\,n\,$ is an odd positive integer.

Sol. Let $\,P(n)\,$ be the mathematical statement i.e., $\,(a^n + b^n)\,$ is divisible by $\,(a+b)\,$ when $\,n=2k+1,\,\,k\in \mathbb N\,\,$ is an odd positive integer.

Now, $\,P(k):\, (a^{2k+1}+b^{2k+1})\,\text{is divisible by }\,(a+b).$ 

So, $\,\,P(1):(a^3+b^3)=(a+b)(a^2-ab+b^2)\,\,$and so $\,P(1)\,\,$ is true. 

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

So, $\,\,a^{2m+1}+b^{2m+1}=r(a+b) \\ \Rightarrow a^{2m+1}=r(a+b)-b^{2m+1}\cdots (1)$

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

which is divisible by $\,\,(a+b).$

and so from (2), we can deduce that  $\,\,P(m+1)\,$ 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 $\,n \in \mathbb N.$

4. If $\,\,n\in \mathbb N\,\,\,$  and  $\,\,(2.1+1)+(2+2+1)+(2+3+1)+... \\+ (2. n + 1) =n^2 + 2n+5\,\,$  

 is true for $\,\,n= m,\,\,$  then prove that it is also true for $\,\,n=m+1.\,$ 

Can we conclude that it is true for all $\,\,n\in \mathbb N\,\,\,$?

Sol. Suppose $\,\,P(m)\,$ is true. 

So,  $\,P(m):(2.1+1)+(2+2+1)+(2+3+1)+... + (2. m + 1) \\=m^2 + 2m+5 $ 

And so, $\,\,P(m+1):(2.1+1)+(2+2+1)+(2+3+1)+... + (2. \overline{m+1} + 1) \\=m^2 + 2m+5+2m+3\\=(m+1)^2 + 2(m+1)+5 $

Hence, the statement is true for $\,\,n=m+1.$

But, $\,\,P(1):2.1+1 \neq 1^2+2.1+5$ and hence $\,\,P(1)\,$ is not true. 

Finally, we can say the statement is not true for all $\,\,n \in \mathbb N.$

5. Prove by induction method that, $\,n(n^2-1)\,$ is divisible by $\,24\,$ when $\,n\,$ is an odd positive integer.

Sol.  Since $\,\,n\,$ is an odd positive integer, let $\,n=2k-1,\,\,k \in \mathbb N.$

So, $\,\,n(n^2-1)\\=(2k-1)\{(2k-1)^2-1\}\\=(2k-1)(2k-1+1)(2k-1-1)\\=(2k-1)(2k)(2k-2)\\=4k(k-1)(2k-1)$

Let the mathematical statement be denoted by $\,P(k),\,$ i.e., $\,\,P(K):4k(k-1)(2k-1)\,\,$ is always divisible by $\,24.\,$

So, $\,\,P(1): 0,\,$which is always divisible by $\,24.$ So, $\,P(1)\,$ is true.

Let $\,P(m)\,$ is true. 

So, $\,\,P(m): 4m(m-1)(2m-1)=24r \\ \Rightarrow 2m^3=6r+3m^2-m \cdots(1)$

Now, $\,\,P(m+1): 4(m+1)m(2m+1)\\=4(2m^3+3m^2+m)\\=4(6r+3m^2-m+3m^2+m) \,\,[\text{By (1)}]\\=24(r+m^2),$

which is divisible by $\,24\,\,$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 $\,\,P(k)\,$ is true for all $\,k \in \mathbb N.$

6. Find the positive integer $\,n\,$ for which the inequality $\,\,2^n>n^2.$

Sol.  Let $\,\,P(n): 2^n>n^2 \\ P(1): 2>1 ,\,\,\text{True} \\ P(2): 4>4,\,\,\text{Not True}\\ P(3): 8>9\,\,\text{Not True} \\ P(4): 16 >16\,\,\text{Not True} \\ P(5): 32 >25\,\,\text{True} \\ P(6): 64>36\,\,\text{True} .$

Let us assume, $\,\,P(n)\,$ is true for $\,\,n \geq 5.$

Clearly, $\,\,P(5)\,\,$ is true.

Let $\,\,P(m)\,\,$ is true. Then, $\,\,2^m>m^2.$

Now, $\,\, 2^m >m^2\,\,\text{and}\,\,2^{m}>2m+1 \\ \therefore 2^m+2^m >m^2+2m+1 \\ \therefore 2^{m+1}>(m+1)^2.$

Since, $\,\,P(5)\,\,$ 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 $\,n \geq 5.$

7. Prove by the method of mathematical induction that for all $\,\,n \in \mathbb N,\,\,\,3^{2n}$  when divided by $\,8,\,$ the remainder is always $\,1.\,$

Sol. $\,\,P(1): 3^{2.1}=9,\,\,$ which when divided by $\,8\,$ gives $\,1\,$ as a remainder.

Let $\,P(m)\,$ be true. 

So, $\,\,3^{2m}=8k+1.$

Now, $\,\,P(m+1): 3^{2(m+1)}=9.3^{2m}\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=9(8k+1)\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=72k+9\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=8(9k+1)+1,\,$

which when divided by $\,8,\,$ gives $\,1\,$ as a remainder.

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 $\,n \in \mathbb N.$

8. Prove by induction that $\,\,5^{n+1}+4.6^{n}\,\,$ when divided by $\,\,20\,\,$ leaves the same remainder $\,9\,$ for all $\,n \in \mathbb N$.

Sol. $\,\,P(1): 5^{1+1}+4.6^1=49,\,\,$ which when divided by $\,20\,$ leaves $\,9\,$ as a remainder.

Let $\,P(m)\,$ be true. 

So, $\,\,5^{m+1}+4.6^m\\=20k+9 \\ \Rightarrow 5^{m+1}=20k+9-4.6^m \cdots(1)$

Now, $\,\,P(m+1): 5^{m+1+1}+4.6^{m+1}\\=5(20k+9-4.6^m)+24.6^m\,\,[\text{By (1)}]\\=100k+45+4.6^m\\=100k+45+4.(1+5)^m\\=100k+45+4.(1+^mC_1.5+^mC_2.5^2\\+\cdots+^mC_m.5^m)\\=100k+49+20(^mC_1+^mC_2.5\\+\cdots+^mC_m.5^{m-1})\\=20.(5k+2+^mC_1+^mC_2.5\\+\cdots+^mC_m.5^{m-1})+9,$

which when divided by $\,20,\,$ leaves $\,9\,$ as a remainder.

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 $\,n \in \mathbb N.$

9. Prove by induction that $\,8.7^n + 4^{n+2}\,$ is divisible by $\,24\,$ but not by $\,48\,$ for all $\,n\in \mathbb N.$

Sol. Let $\,P(n)\,$ be $\,8.7^n + 4^{n+2}\,$ is divisible by $\,24\,$ but not by $\,48\,$ for all $\,n\in \mathbb N.$

Now, $\,P(1): 8.7^1+4^{1+2}=120\,$ which is divisible by $\,24.$

So, $\,P(1)\,$ is true. 

Let $\,P(m)\,$ be true and so, $\,8.7^m + 4^{m+2}=24k,\,k \in \mathbb N \\ \Rightarrow4^{m+2}=24k-8.7^m$

Now, $\,P(m+1): 8.7^{m+1}+4^{m+3}\\=56.7^m+4.4^{m+2}\\=56.7^m+4(24k-8.7^m)\\=24.7^m+4.24k\\=24(7^m+4k),\,\,\text{which is divisible by}\,\,24.$

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 $\,n \in \mathbb N.$

Now, let $\,\,P_1(n)\,\,$ be another mathematical statement where, $\,P_1(n)\,:8.7^n+4^{n+2}\,\,$ is always divisible by $\,48.$

But, $\,P_1(1): 8.7^1+4^{1+2}=120\,$ which is not divisible by $\,48\,\,$i.e.,  $\,P_1(1)\,$ is not true. Hence , follows the second result.

10. Prove by induction that $\,2^{2^n} +1\,$ has $\,7\,$ in unit's place for all $\,\,n\geq2.$

Sol. Let $\,\,P(n): 2^{2^n} +1\,$ has $\,7\,$ in unit's place for all $\, n \geq2.$

Now , $\,\,P(2): 2^{2^2}+1=16+1=17\,$ and so, $\,P(1)\,$ is true. 

Let $\,\,P(m): 2^{2^m} +1\,$ has $\,7\,$ in unit's place for all $\, m\geq2.$

and so, $\,\,P(m): 2^{2^m}+1=10p+7,\,\,p \in \mathbb N \\ \therefore P(m+1):2^{2^{m+1}}+1 \\=(2^{2^m})^2+1 \\=(10p+7-1)^2+1\\=(10p+6)^2+1\\=100p^2+120p+36+1\\=100p^2+120p+37$

Now , since $\,P(2)\,$ 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 $\,n \geq 2.$

11. Using mathematical induction prove that for every integer $\,n,\,\, |\sin{nx}|\leq| \sin x| $.

Sol.  $\,\,P(1): |\sin(1.x)| \leq 1.|\sin x|.$

Hence, $\,P(1)\,$ is true.  Let $\,\,P(m)\,$ is true.

Now, $\,\,P(m+1): |\sin(m+1)x|\\=|\sin(mx+x)|\\=|\sin mx \cos x+\cos mx \sin x|\\ \leq|\sin(mx)|.1+1.|\sin x|\\ \leq  m|\sin x|+|\sin x|\\ \leq (m+1)|\sin x| $

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 $\,n \in \mathbb N.$