Differentiation (Part-38) | S N De

 

$\,1(i)\,\,$ If $\,S\,$ be the sum, $\,P\,$ be the product and $\,R\,$ the sum of the reciprocals of $\,n\,$ terms in a G.P. , prove that, $\,P^2=(\frac SR)^n$

Sol. $\,S=a+ar+ar^2+\cdots+ar^{n-1}\\ \Rightarrow S= a\cdot \frac{r^n-1}{r-1},\\ P=a .(ar).(ar^2).\cdots(ar^{n-1})\\~~~=a^nr^{1+2+3+\cdots+(n-1)}\\~~~=ar^{\frac{n(n-1)}{2}}, \\ R=\frac 1a+\frac{1}{ar}+\frac{1}{ar^2}+\cdots+\frac{1}{ar^{n-1}}\\~~~=\frac 1a.\frac{\frac{1}{r^n}-1}{\frac 1r-1}\\~~~=\frac 1a.\frac{r^n-1}{r-1}.\frac{1}{r^{n-1}}\\ \therefore \frac SR=a^2 \cdot r^{n-1} \\ \Rightarrow (\frac SR)^n=a^{2n}.r^{n(n-1)}\\~~~~~~~~~~~~~~=[a^n r^{\frac{n(n-1)}{2}}]^2=P^2\,\,\text{(proved)}$

$\,1(ii)\,$ If $\,S_1,S_2,S_3\,$ be respectively the sums of $\,n,2n\,$ and $\,3n\,$ terms of a G.P., prove that, $\,S_1(S_3-S_2)=(S_2-S_1)^2.$

Sol. $\,S_3-S_2\\=a.\frac{r^{3n}-1}{r-1}-a.\frac{r^{2n}-1}{r-1}\\=\frac{a}{r-1}[(r^{3n}-1)-(r^{2n}-1)]\\=\frac{a}{r-1}(r^{3n}-r^{2n}) \rightarrow(1)\\ \Rightarrow S_1(S_3-S_2)\\=a.\frac{r^n-1}{r-1} \times \frac{a}{r-1} \cdot r^{2n}(r^n-1)\,\,\text{[By (1)]}\\=\frac{a^2}{(r-1)^2}\times r^{2n}(r^n-1)^2\rightarrow(2)$

Again, $\,S_2-S_1\\=a \cdot \frac{r^{2n}-1}{r-1}-a\cdot \frac{r^n-1}{r-1}\\=\frac{a}{r-1} \cdot [(r^{2n}-1)-(r^n-1)]\\=\frac{a}{r-1}\cdot (r^{2n}-r^n)\\=\frac{a}{r-1}\cdot r^n(r^n-1)\\ \Rightarrow (S_2-S_1)^2=\frac{a^2}{(r-1)^2}\times r^{2n}(r^n-1)^2\rightarrow(3)$

From $\,(2)\,$ and $\,(3),\,$ we can conclude that $\,S_1(S_3-S_2)=(S_2-S_1)^2.$

$\,2.\,$ If the  sum of first $\,n\,$ terms of a G.P. is $\,S\,$ and the sum of its first $\,2n\,$ terms is $\,5S\,$ , then show that the sum of its first $\,3n\,$ terms is $\,21S.$

Sol. According to the problem,

$\,S=a.\frac{r^n-1}{r-1}\rightarrow(1),\\ 5S=a.\frac{r^{2n}-1}{r-1} \\ \Rightarrow 5S=a.\frac{(r^n)^2-1}{r-1}=a.\frac{(r^n+1)(r^n-1)}{r-1} \\ \Rightarrow 5S=(r^n+1).[a.\frac{r^n-1}{r-1}] \\ \Rightarrow 5S=(r^n+1)S\,\,~~\text{[By (1)]} \\ \Rightarrow 5=r^n+1 \\ \Rightarrow r^n=5-1 \\ \Rightarrow r^n=4\rightarrow(2)$

Then, from $\,(1),(2)\,$ we get, 

$\,S=\frac{a}{r-1}. (4-1) \\ \Rightarrow \frac S3=\frac{a}{r-1}\rightarrow(3)$

Now, the sum of its first $\,3n\,$ terms is 

$= a.\frac{r^{3n}-1}{r-1}\\=a.\frac{(r^n)^3-1}{r-1}\\=\frac{a}{r-1}.[4^3-1]\\=\frac S3 \times 63\\=21S\,\,\text{(showed)}$

$\,3.\,$ If $\,S_n=1+\frac 12+\frac{1}{2^2}+\cdots+\frac{1}{2^{n-1}},\,\,$ find the least value of $\,n\,$ for which $\,2-S_n<\frac{1}{100}.$

Sol.  $\,2-S_n<\frac{1}{100}\,\,\text{(Given)}\\ \Rightarrow 2-1.\frac{1-\frac{1}{2^n}}{1-\frac 12}<\frac{1}{100} \\ \Rightarrow 2-2\left(1-\frac{1}{2^n}\right)<\frac{1}{100} \\ \Rightarrow 2-2+\frac{2}{2^n}<\frac{1}{100} \\ \Rightarrow 2^n>200 \rightarrow(1)$

From $\,(1),\,$ we see that $\,2^7<200<2^8\,\,$ and so we can conclude that the the least value of $\,n\,$ is $\,8,\,$ for which $\,2-S_n<\frac{1}{100}.$

$\,4.\,$ If $\,\,u_1,u_2,u_3,\cdots\,$ form a G.P. with common ratio $\,k,\,$ find the sum of $\,u_1u_2+u_2u_3+\cdots+u_nu_{n+1}\,\,$ in terms of $\,u_1\,$ and $\,k.$

Sol.  Since $\,\,u_1,u_2,u_3,\cdots\,$ form a G.P. with common ratio $\,k,\,$

$\,u_2=u_1k,\\u_3=u_2k=(u_1k)k=u_1k^2,\\ u_4=u_3k=(u_1k^2)k=u_1k^3, \\ \vdots \\ u_n=u_{1}k^{n-1},\\u_{n+1}=u_1k^n.$

So, $\,u_1u_2+u_2u_3+\cdots+u_nu_{n+1}\\=u_1(u_1k)+(u_1k)(u_1k^2)+\cdots\\ +(u_{1}k^{n-1})(u_1k^n)\\=u_1^2k[1+k^2+k^4+\cdots +k^{2n-2}]\\=u_1^2k [1+k^2+(k^2)^2+\cdots +(k^2)^{n-1}]\\=u_1k.\left[\frac{(k^2)^n-1}{k^2-1}\right]\\=u_1k \cdot \frac{k^{2n}-1}{k^2-1}\,\,\text{(ans.)}$

$\,5.\,$ Find the sum to $\,n\,$ terms of the following series:

$\,\left(x+\frac 1x\right)^2+\left(x^2+\frac{1}{x^2}\right)^2+\left(x^3+\frac{1}{x^3}\right)^2\\+\left(x^4+\frac{1}{x^4}\right)^2+\cdots $

Sol. $\,\left(x+\frac 1x\right)^2+\left(x^2+\frac{1}{x^2}\right)^2+\left(x^3+\frac{1}{x^3}\right)^2\\+\left(x^4+\frac{1}{x^4}\right)^2+\cdots \\=[x^2+\frac{1}{x^2}+2.x.\frac 1x]+[x^4+\frac{1}{x^4}+2.x^2.\frac{1}{x^2}]\\+[x^6+\frac{1}{x^6}+2.x^3.\frac{1}{x^3}]+[x^8+\frac{1}{x^8}+2.x^4.\frac{1}{x^4}]\\+\cdots\\=(x^2+x^4+x^6+\cdots\text{to n terms })\\+(2+2+2+\cdots \text{to n terms})\\+\left(\frac{1}{x^2}+\frac{1}{x^4}+\frac{1}{x^6}+\cdots \text{to n terms}\right)\\=x^2.\frac{(x^2)^n-1}{x^2-1}+2n+\frac{1}{x^2} \cdot \frac{1-\frac{1}{x^{2n}}}{1-\frac{1}{x^2}}\\=x^2 \cdot\frac{x^{2n}-1}{x^2-1}+2n+\frac{1}{x^2-1} \cdot \frac{x^{2n}-1}{x^{2n}}\\=x^2 \cdot\frac{x^{2n}-1}{x^2-1}+2n+\frac{x^{2n}-1}{x^2-1} \cdot \frac{1}{x^{2n}}\\=\frac{x^{2n}-1}{x^2-1}(x^2+x^{-2n})+2n\,\,\text{(ans.)}$ 

$\,6.\,$ If the first term, common ratio and the sum of first $\,n\,$ terms of a G.P. be $\,a,r\,$ and $\,S_n\,$ respectively, find the value of $\,S_1+S_2+S_3+\cdots +S_n.$

Sol. We know, $\,S_n=a. \frac{r^n-1}{r-1} , \\ S_1=a.\frac{r-1}{r-1},\\ S_2=a.\frac{r^2-1}{r-1},\\ S_3=a.\frac{r^3-1}{r-1},\\ \vdots\\ \therefore \,\, S_1+S_2+S_3+\cdots +S_n\\=\frac{a}{r-1}[(r-1)+(r^2-1)+(r^3-1)+\cdots+(r^n-1)]\\=\frac{a}{r-1}[(r+r^2+r^3+\cdots +r^n)-n]\\=\frac{a}{r-1}.\frac{r(r^n-1)}{r-1}-\frac{an}{r-1}\\=\frac{ar(r^n-1)}{(r-1)^2}-\frac{an}{r-1}\,\,\text{(ans.)}$

$\,7.\,$  When a certain golf-ball is dropped on a piece of pavement, it bounces to a height of three-fifth the height from which it fell. If the ball is dropped from a height of $\,100\,$ cm, how far has it travelled when it hits the ground for the tenth time? How high will it rise on the next bounce?

Sol. For the first time, the golf ball is dropped from a height of $\,100\,$ cm. 

At the second time, the total distance covered by the ball 

$=100+2 \cdot 100 \cdot \frac 35\,$ cm.

At the third time, the total distance covered by the ball 

$=100+2 \cdot 100 \cdot \frac 35+2 \cdot 100 \cdot (\frac 35)^2\,$ cm.

Similarly, at the $\,10\,$-th time, the total distance covered by the ball 

$=100+2 \cdot 100 \cdot\left[ \frac 35+ (\frac 35)^2\cdots +(\frac 35)^9\right]\,\,\text{cm.}\\=100+2\cdot 100 \cdot \frac 35 \cdot \frac{1-(\frac 35)^9}{1-\frac 35}\,\,\text{cm.}\\=100+2\cdot 100 \cdot \frac 35 \cdot \frac 52 \cdot \left[1-(\frac 35)^9\right]\,\,\text{cm.}\\=100+296.97\,\,\text{cm.}\\=396.97\,\,\text{cm. (approx)}$ 

Total distance  travelled when it hits the ground for the tenth time is :

$=100 \times \left(\frac 35\right)^10\\=0.6047\,\,\text{cm. (approx)}$ 

$\,8.\,$ If $\,S_1,S_2,S_3,\cdots, S_n\,$ be the sums of first $\,n\,$ terms of $\,n\,$ G.P.'s whose first terms are each unity and the common ratios are $\,1,2,3,\cdots ,n\,$ respectively, prove that, 

$\,S_1+S_2+2S_3+3S_4+\cdots+(n-1)S_n=1^n+2^n+3^n+\cdots+n^n.$

Sol. Here, $\,S_1=1+1+1+\cdots \text{to n terms}=n, \\ S_2=1+2+2^2+\cdots \text{to n terms}=1.\frac{2^n-1}{2-1},\\ S_3=1+3+3^2+\cdots \text{to n terms}=1.\frac{3^n-1}{3-1},\\ S_4=1+4+4^2+\cdots \text{to n terms}=1.\frac{4^n-1}{4-1},\\ \vdots \\ S_n=1+n+n^2+\cdots \text{to n terms}=1.\frac{n^n-1}{n-1}.$

Hence, $\,S_1+S_2+2S_3+3S_4+\cdots+(n-1)S_n\\=n+(2^n-1)+2.\frac{3^n-1}{3-1}+3.\frac{4^n-1}{4-1}\\+\cdots +(n-1).\frac{n^n-1}{n-1}\\=n+(2^n-1)+(3^n-1)+(4^n-1)\\+\cdots +(n^n-1)\\=n+ 2^n+3^n+4^n+\cdots +n^n-(n-1)\\=1^n+2^n+3^n+\cdots+n^n\,\,\text{(proved)}$

$\,9.\,$ If the A.M. of two numbers be twice their G.M. , prove that two numbers are in the ratio $\,\,\,(2+\sqrt 3): (2-\sqrt 3).$

Sol. Let the two numbers be $\,a,b,\,\,$ so that 

$\,\frac{a+b}{2}=2\sqrt{ab} \\ \Rightarrow \left(\frac{a+b}{2}\right)^2=\left(2\sqrt{ab}\right)^2 \\ \Rightarrow (a+b)^2=4 \times 4ab \\ \Rightarrow a^2+2ab+b^2=16ab \\ \Rightarrow a^2-14ab+b^2=0 \\ \Rightarrow (\frac ab)^2-14(\frac ab)+1=0 \\ \therefore \frac{a}{b}=\frac{14 \pm\sqrt{14^2-4\times 1\times 1}}{2\times 1}\\~~~~~~~=7\pm 4\sqrt 3\,\,; \\ \text{so,}\,\, \frac ab=7 +4\sqrt3\\~~~~~~~=(2+\sqrt3)^2\\~~~~~~~=\frac{(2+\sqrt3)^2}{(2+\sqrt3)(2-\sqrt3)}\\~~~~~~~=\frac{2+\sqrt3}{2-\sqrt3}\rightarrow(1), \\ \text{Similarly},\,\,\\~~~~ \frac ab=7-4\sqrt3\\~~~~~~~=(2-\sqrt3)^2\\~~~~~~~=\frac{(2-\sqrt3)^2}{(2+\sqrt3)(2-\sqrt3)}\\~~~~~~~=\frac{2-\sqrt3}{2+\sqrt3}.$

Hence, from $\,(1),\,$ we can say that two numbers are in the ratio $\,\,\,(2+\sqrt 3): (2-\sqrt 3).$

$\,10.\,$ If $\,A\,$ and $\,G\,$ be the A.M. and G.M.  respectively of positive numbers , show that the numbers are $\,A \pm \sqrt{A^2-G^2}.$

Sol. For two numbers $\,a,\,b$, we have,

$\,A=\frac{a+b}{2},\,\,G=\sqrt{ab}.$

Now, $\,G^2=ab \Rightarrow b=\frac{G^2}{a}.$

Again, $\,A=\frac 12\times (a+b)\\~~~=\frac 12\cdot \left(a+\frac{G^2}{a}\right)\\ \Rightarrow 2A=a+\frac{G^2}{a} \\ \Rightarrow 2Aa=a^2+G^2 \\ \Rightarrow a^2-2Aa+G^2=0 \\ \Rightarrow a=\frac{-(-2A)\pm\sqrt{(-2A)^2-4 \times 1\times G^2}}{2\times 1}\\~~~~~~~=A \pm \sqrt{A^2-G^2}\,\,\text{(showed)}$