Differentiation (Part-38) | S N De

 In the previous articles, we have discussed Short and Long Answer Type Questions. Now, In this article , we will discuss Very Short Answer Type Questions and their solutions. So, let's start.

EXERCISE-3

1. If $\,\,(2^{2n}-1)\,$ is divisible by $\,3,\,$ then show that, $\,[2^{2(n+1)}-1]\,$ is also divisible by $\,\,3.$

Sol. $\,[2^{2(n+1)}-1]\\=[2^{2n}.2^2-1]\\=4.2^{2n}-1\\=3.2^{2n}+(2^{2n}-1)\\=3.2^{2n}+3l,[\text{Since,}\,\,\,(2^{2n}-1)\, \text{is divisible by} \,\,3,\\(2^{2n}-1)=3l ]\\=3(2^{2n}+l)\\=3l_1\,\,[l_1=2^{2n}+l (\in \mathbb I)]$ 

Hence, $\,[2^{2(n+1)}-1]\,$ is also divisible by $\,\,3.$

2. If for all $\,\,n \in \mathbb N,\,\,2n+7 <(n+3)^2,\,\,$ then show that, $\,\,2(n+1)+7<(n+4)^2.$

Sol. Since for all $\,\,n \in \mathbb N,\,\,2n+7 <(n+3)^2,\,\,$ we replace $\,n\,$ with $\,(n+1),\,$ and get $\,\,2(n+1)+7<(n+1+3)^2 \Rightarrow 2(n+1)+7 <(n+4)^2.$ 

3. If $\,(2^{3n}-1)\,$ is divisible by $\,7,\,$then prove that , $\,\,[2^{3(n+1)}-1]\,$ is also divisible by $\,7.$

Sol. By question, $\,(2^{3n}-1)=7l, \,\,l(\neq 0)=\text{Constant} \in \mathbb Z \cdots (1)$

Now, $\,2^{3(n+1)}-1\\=2^{3n}.2^3-1\\=8.2^{3n}-1\\=2^{3n}-1+7.2^{3n}\\=7l+7.2^{3n}\,\,[\text{By (1)}]\\=7(l+2^{3n})\\=7l_1,\,\,\text{where,}\,\,\,\, l+2^{3n}=l_1 \in \mathbb Z.$

Hence , $\,\,[2^{3(n+1)}-1]\,$ is also divisible by $\,7.$

4. If $\,\,1.1!+2.2!+3.3! +\cdots+n.n!=(n+1)!-1,\,\,$ show that $\,\,1.1!+2.2!+3.3! +\cdots\\+n.n!+(n+1).(n+1)!=(n+2)!-1$

Sol. $\,\,1.1!+2.2!+3.3! +\cdots\\+n.n!+(n+1).(n+1)!\\=(n+1)!-1+(n+1).(n+1)!\\=(n+1)!(n+1+1)-1\\=(n+2)(n+1)!-1\\=(n+2)!-1$

5. If $\,(10^{2n-1}+1)\,$is divisible by $\,11\,$, then prove that $\,(10^{2n+1}+1)\,$ is also divisible by $\,11$.

Sol. Since, $\,(10^{2n-1}+1)\,$is divisible by $\,11 \Rightarrow 10^{2n-1}+1=11p,\,\text{where,}\,\,p \,\,\text{is constant}\, \in \mathbb Z.$

Now, $\,\,10^{2n+1}+1\\=10^{2n-1}.10^2+1\\=10^{2n-1}+1+99.10^{2n-1}\\=10^{2n-1}+1+11.9.10^{2n-1}\\=11p+11.9.10^{2n-1}\\=11(p+9.10^{2n-1})\\=11p_1,\,\,\text{where,}\,\,p_1=p+9.10^{2n-1} \in \mathbb Z$

Hence, follows the result.

6. If $\,(15^{2n-1}+1)\,$is divisible by $\,16\,$, then prove that $\,(15^{2n+1}+1)\,$ is also divisible by $\,16$.

Sol. Since, $\,(15^{2n-1}+1)\,$is divisible by $\,16 \Rightarrow 15^{2n-1}+1=16p,\,\text{where,}\,\,p \,\,\text{is constant}\, \in \mathbb Z.$

Now, $\,\,15^{2n+1}+1\\=15^{2n-1}.15^2+1\\=15^{2n-1}.225+1\\=15^{2n-1}+1+224.15^{2n-1}\\=16p+224.15^{2n-1}\\=16(p+14.15^{2n-1})\\=16p_1,\,\,\text{where,}\,\,p_1=p+14.15^{2n-1}\in \mathbb Z$

Hence, follows the result.

7. If $\,\,[12^{n}+25^{n-1}]\,\,$ is divisible by $\,13,\,$ then show that,$\,\,[12^{n+1}+25^{n}]\,\,$is also divisible by $\,13$.

Sol. By question, $\,\,[12^{n}+25^{n-1}]=13p\,\text{=constant} \in\mathbb Z$ 

Now, $\,\,[12^{n+1}+25^{n}]\\=12.12^n+25.25^{n-1}\\=12(12^n+25^{n-1})+13.25^{n-1}\\=12.13p+13.25^{n-1}\\=13(12p+25^{n-1})\\=13p_1,\,\,\text{where,}\,\,12p+25^{n-1}=p_1 \in \mathbb Z$

Hence, follows the result.

8. If $\,[n^3+ (n+1)^3 + (n+2)^3]\,\,$ is always divisible by $\,9,\,$ them show that, $\,[(n+1)^3+ (n+2)^3 + (n+3)^3]\,\, $is also divisible by $\,9$. 

Sol. By question, $\,[n^3+ (n+1)^3 + (n+2)^3]=9p,\,\,p \in \mathbb Z \\ \Rightarrow (n+1)^3 + (n+2)^3=9p-n^3 \cdots(1)$

Now,  $\,[(n+1)^3+ (n+2)^3 + (n+3)^3]\\=9p+(n+3)^3-n^3\,\,[\text{By (1)}]\\=9p+(n+3-n)\left((n+3)^2+n(n+3)+n^2\right)\\=9p+3(n^2+6n+9+n^2+3n+n^2)\\=9p+3(3n^2+9n+9)\\=9(p+n^2+3n+3)\\=9p_1, \,\,\text{where,}\,\,p+n^2+3n+3=p_1 \in \mathbb Z$

Hence, follows the result.

9. If $\,\,n \in \mathbb N,\,x >-1 \,\,\text{and}\,\,(1+x)^n \geq 1+nx,\,$ then prove that, $\,(1+x)^{n+1} \geq 1+(n+1)x$

Sol. $\,\,(1+x)^{n+1} =(1+x)^n.(1+x) \\ ~~~~~~~~~~~~~~~~~~~\geq (1+nx)(1+x) \\~~~~~~~~~~~~~~~~~~~=1+x+nx+nx^2\\~~~~~~~~~~~~~~~~~~~=1+(n+1)x+nx^2 \\~~~~~~~~~~~~~~~~~~~~\geq 1+(n+1)x\,\,[x >-1, n \in \mathbb N \Rightarrow nx^2 \geq 0]$

10. Let $\,f(n) = n(n+1)(2n+1);\,\,$ if $f(n)\,$ is always divisible by $\,6,\,$ then prove that, $\,f(n+1)\,$ is also divisible by $\,6.$

Sol. $\,f(n) = n(n+1)(2n+1)=6p,\,\,p\in \mathbb Z \\ f(n+1)=(n+1)(n+2)(2n+3)\\~~~~~~~~~~~~~~=(n+2)(n+1)[(2n+1)+2]\\~~~~~~~~~~~~~~=n(n+1)[(2n+1)+2]+2(n+1)[(2n+1)+2]\\~~~~~~~~~~~~~~=n(n+1)(2n+1)+2n(n+1)+2(n+1)(2n+3)\\~~~~~~~~~~~~~~=6p+2(n+1)(n+2n+3)\\~~~~~~~~~~~~~~=6p+2(n+1)\times 3(n+1)\\~~~~~~~~~~~~~~=6p+6(n+1)(n+1)\\~~~~~~~~~~~~~~=6[p+(n+1)^2]\\~~~~~~~~~~~~~~=6p_1,\,\,\text{where,}\,\,p+(n+1)^2=p_1 \in \mathbb Z$

Hence,$\,\,\,f(n+1)\,$ is also divisible by $\,6.$

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