Differentiation (Part-38) | S N De

 

$\,9.\,$ In an infinite geometric G.P., each term is equal to three times the sum of all the terms that follow it and the sum of first two terms is $\,15;$ find the sum of the series to infinity.

Sol. Let us consider the infinite geometric series $\,a,ar,ar^2,ar^3,\cdots ;$ where $\,a\,$ is the first term and $\,r\,$ being the common ratio.

By the condition, we have

$\,a=3(ar+ar^2+ar^3+\cdots)\\~~~=3 \cdot \frac{ar}{1-r}\\ \Rightarrow 1=\frac{3r}{1-r}\\ \Rightarrow 1-r=3r \\ \Rightarrow 1=4r \\ \Rightarrow r=\frac 14.$

Now, since the sum of first two terms is $\,15,$

so, $\,a+ar=15\\ \Rightarrow a(1+r)=15 \\ \Rightarrow a\left(1+\frac 14\right)=15 \\ \Rightarrow a \times \frac 54=15 \\ \Rightarrow a= 15 \times \frac 45=12.$

Hence, $\,a+ar+ar^2+\cdots\\=\frac{a}{1-r}\\=\frac{12}{1-\frac 14}\\=\frac{12}{3/4}\\=\frac{12 \times 4}{3}\\=4 \times 4\\=16\,\,\,\text{(ans.)}$

$\,10.\,$ If $\,x=a+\frac ar+\frac{a}{r^2}+\cdots \infty\,\,\left(-1<\frac 1r<1\right),\\y=b-\frac br+\frac{b}{r^2}-\cdots \infty,\\z=c+\frac{c}{r^2}+\frac{c}{r^4}+\cdots \infty,\,\,$ show that , $\,\,\frac{xy}{z}=\frac{ab}{c}.$

Sol. We have, $x=a+\frac ar+\frac{a}{r^2}+\cdots \infty\\~~=\frac{a}{1-\frac 1r}\,\,[*]\\~~=\frac{ar}{r-1}\rightarrow(1)\\y=b-\frac br+\frac{b}{r^2}-\cdots \infty\\~~=\frac{b}{1-(-1/r)}~~[**]\\~~=\frac{br}{r+1}\rightarrow(2)$

Note[*]: $\,x\,$ is an infinite geometric series with first term $\,a,\,$and common ratio $=\frac 1r.$

Note[**]: $\,y\,$ is an infinite geometric series with first term $\,b,\,$and common ratio $=-\frac 1r.$

Similarly, $\,z=c+\frac{c}{r^2}+\frac{c}{r^4}+\cdots \infty\\~~=\frac{c}{1-\frac {1}{r^2}}\\~~=\frac{cr^2}{r^2-1}\rightarrow(3)$

Hence, from $\,(1),\,(2),\,(3)\,$ we get,

$\,~~~\frac{xy}{z}\\=\frac{\frac{ar}{r-1} \times \frac{br}{r+1}}{\frac{cr^2}{r^2-1}}\\=\frac{\frac{abr^2}{r^2-1}}{\frac{cr^2}{r^2-1}}\\=\frac{ab}{c}\,\,\text{(proved)}$

$\,11.\,$ If $\,|a|<1,\,|b|<1,\,$ then show that

$\,a(a+b)+a^2(a^2+b^2)+a^3(a^3+b^3)+\cdots=\frac{a^2}{1-a^2}+\frac{ab}{1-ab}$

Sol. $a(a+b)+a^2(a^2+b^2)+a^3(a^3+b^3)+\cdots\\=(a^2+a^4+a^6+\cdots)+(ab+a^2b^2+a^3b^3+\cdots)\\=S_1+S_2\rightarrow(1)$

Clearly, $\,S_1\,$ represents an infinite geometric series with first term $=a^2,\,$ common ratio $=\frac{a^4}{a^2}=a^2.$

So, $\,S_1=a^2+a^4+a^6+\cdots=\frac{a^2}{1-a^2}\rightarrow(2)$

Similarly, $\,S_2\,$ represents an infinite geometric series with first term $=ab,\,$ common ratio $=\frac{a^2b^2}{ab}=ab.$

So, $\,S_2=ab+a^2b^2+a^3b^3+\cdots=\frac{ab}{1-ab}\rightarrow(3)$

Hence, from $(1),\,(2),\,(3)\,$ we get the desired result.

$\,12.\,$ If for real numbers $\,\,a,b,c,r\,\,$ with $\,|r|>1,\,x=a+\frac ar+\frac{a}{r^2}+\cdots,~~~y=b-\frac br+\frac{b}{r^2}-\frac{b}{r^3}+\cdots$ and $\,z=c+\frac{c}{r^2}+\frac{c}{r^4}+\cdots,$then show that $\,xyc=abz.$

Sol. We have, $x=a+\frac ar+\frac{a}{r^2}+\cdots \infty\\~~=\frac{a}{1-\frac 1r}\,\,[*]\\~~=\frac{ar}{r-1}\rightarrow(1)\\y=b-\frac br+\frac{b}{r^2}-\cdots \infty\\~~=\frac{b}{1-(-1/r)}~~[**]\\~~=\frac{br}{r+1}\rightarrow(2)$

Note[*]: $\,x\,$ is an infinite geometric series with first term $\,a,\,$and common ratio $=\frac 1r.$

Note[**]: $\,y\,$ is an infinite geometric series with first term $\,b,\,$and common ratio $=-\frac 1r.$

Similarly, $\,z=c+\frac{c}{r^2}+\frac{c}{r^4}+\cdots \infty\\~~=\frac{c}{1-\frac {1}{r^2}}\\~~=\frac{cr^2}{r^2-1}\rightarrow(3)$

Hence, from $\,(1),\,(2),\,(3)\,$ we get,

$\,~~~\frac{xy}{z}\\=\frac{\frac{ar}{r-1} \times \frac{br}{r+1}}{\frac{cr^2}{r^2-1}}\\=\frac{\frac{abr^2}{r^2-1}}{\frac{cr^2}{r^2-1}}\\=\frac{ab}{c} \\ \Rightarrow \frac{xy}{z}=\frac{ab}{c}\\ \Rightarrow xyc=abz\,\,\text{(proved)}$

$\,13.\,$ If for two reals $\,a,b\,$ with $\,|a|>1,\,|b|>1,\,x=1+\frac 1a+\frac{1}{a^2}+\cdots,\\ y=1+\frac 1b+\frac{1}{b^2}+\cdots$ 

then show that the sum of the series $\,1+ab+a^2b^2+\cdots=\frac{1-x-y+xy}{1-x-y}.$

Sol. According to the  problem, $\,x\,$ represents an infinite geometric series with first term $=1,\,$ common ratio $=\frac 1a.$

So, $\,x=\frac{1}{1-1/a} \\ \Rightarrow 1-1/a=\frac 1x \\ \Rightarrow 1-\frac 1x=\frac 1a \\ \Rightarrow \frac 1a=\frac{x-1}{x} \\ \Rightarrow a=\frac{x}{x-1}.$

Similarly, $\,y\,$ represents an infinite geometric series with first term $=1,\,$ common ratio $=\frac 1b.$

So, $\,y=\frac{1}{1-1/b} \\ \Rightarrow 1-1/b=\frac 1y \\ \Rightarrow 1-\frac 1y=\frac 1b \\ \Rightarrow \frac 1b=\frac{y-1}{y} \\ \Rightarrow b=\frac{y}{y-1}.$

So, finally, $\,1+ab+a^2b^2+\cdots\\=\frac{1}{1-ab}\\=\frac{1}{1-\frac{x}{x-1}\cdot\frac{y}{y-1}}\\=\frac{(x-1)(y-1)}{(x-1)(y-1)-xy}\\=\frac{xy-x-y+1}{xy-x-y+1-xy}\\=\frac{1-x-y+xy}{1-x-y}\,\,\text{(showed)}.$

$\,14.\,$ For a real number $\,a\,$ with $\,|a|<1,\,$if $\,x=a-a^3+a^5-\cdots,y=1+a^2+a^4+\cdots$and $\,z=\frac 1a+a^3+a^7+\cdots$ then show that $\,a^2z=xy.$

Sol. According to the problem, $\,x\,$ represents an infinite geometric series with the first term $=a,\,$and common ratio $=\frac{-a^3}{a}=-a^2.$

So, $\,x=a-a^3+a^5-\cdots=\frac{a}{1-(-a^2)}=\frac{a}{1+a^2}\rightarrow(1)$

Again,  $\,y\,$ represents an infinite geometric series with the first term $=1,\,$and common ratio $=\frac{a^2}{1}=a^2.$

So, $\,y=1+a^2+a^4+\cdots=\frac{1}{1-a^2}\rightarrow(2)$

Similarly, $\,z\,$ represents an infinite geometric series with the first term $=\frac 1a,\,$and common ratio $=\frac{a^3}{\frac 1a}=a^4.$

So, $\,z=\frac 1a+a^3+a^7+\cdots=\frac{\frac 1a}{1-a^4}=\frac{1}{a(1-a^4)}\rightarrow(3)$

Now, using $\,(1),\,(2),\,(3)\,$ we get,

$\,a^2z=a^2\cdot \frac{1}{a(1-a^4)}=\frac{a}{1-a^4}\rightarrow(4)$

and $\,xy=\frac{a}{1+a^2}\times \frac{1}{1-a^2}=\frac{a}{1-(a^2)^2}=\frac{a}{1-a^4}\rightarrow(5)$

Hence, from $\,(4),\,(5)\,$ the result follows.

$\,15.\,$ For a real number $\,a\,$ with $\,0<a<\frac 12,\,$ if $\,b=1-a+a^2-a^3+\cdots,c=1-b+b^2-b^3+\cdots\,\,$ and $\,d=1+c+c^2+\cdots ,\,$ find the value of $\,(d-a).$

Sol. According to the problem, $\,b\,$ represents an infinite geometric series with the first term $=1,\,$and common ratio $=\frac{-a}{1}=-a.$

So, $\,b=1-a+a^2-a^3+\cdots=\frac{1}{1-(-a)}=\frac{1}{1+a}$

Again,    $\,c\,$ represents an infinite geometric series with the first term $=1,\,$and common ratio $=\frac{-b}{1}=-b.$

So, $~~~~~c=1-b+b^2-b^3+\cdots \\ \Rightarrow c=\frac{1}{1-(-b)}\\ \Rightarrow c=\frac{1}{1+b}=\frac{1}{1+\frac{1}{1+a}}=\frac{1+a}{1+a+1}=\frac{1+a}{2+a}$

Similarly,    $\,d\,$ represents an infinite geometric series with the first term $=1,\,$and common ratio $=\frac{c}{1}=c.$

So, $\,d=1+c+c^2+\cdots=\frac{1}{1-c}\\ \Rightarrow d=\frac{1}{1-\frac{1+a}{2+a}}=\frac{2+a}{2+a-(1+a)}=2+a \\ \therefore d-a=2\,\,\text{(ans.)}$

$\,16.\,$ If, for $\,0<a<\frac 12,\,$ the sum of the series $\,a+a^2+a^3+\cdots\,$ is $\,b\,$,then show that the series $\,b-b^2+b^3-b^4+\cdots\,$ is also convergent and hence find its sum.

Sol. Let $\,S=a+a^2+a^3+\cdots$

Clearly, $\,S\,$ represents an infinite geometric series with the first term $=a,\,$and common ratio $=\frac{a^2}{a}=a.$

So, $\,S=\frac{a}{1-a}\\ \Rightarrow b=\frac{a}{1-a} \rightarrow(1)$

Now, $\,S_1=b-b^2+b^3-b^4+\cdots$

Clearly, $\,S_1\,$ is an infinite geometric series with the first term $=b,\,$and common ratio $=\frac{-b^2}{b}=-b.$

Hence, $\,S_1=\frac{b}{1-(-b)}=\frac{b}{1+b}\\ \Rightarrow S_1=\frac{a/(1-a)}{1+a/(1-a)}~~~[\text{By (1)}]\\ \Rightarrow S_1=\frac{a}{1-a+a}=a\,\,\text{(ans.)}$

Again, from $\,(1),\,$ we get,

$\,\frac 1b=\frac{1-a}{a} =\frac 1a-1\\ \Rightarrow \frac 1a=1+\frac 1b=\frac{b+1}{b}\\ \Rightarrow a=\frac{b}{b+1}$

Now, since $\,0<a<\frac 12 \\ \Rightarrow 0 < \frac{b}{b+1}<\frac 12 \\ \Rightarrow \frac{b}{b+1}<\frac 12 \\ \Rightarrow 2b<1+b \\ \Rightarrow 2b-b<1 \\ \Rightarrow b <1 \rightarrow(2)$ 

So, by $\,(2),\,$ we can say that the given series is convergent and its sum is $\,a.$

$\,17.\,$ If $\,x=\sum_{n=0}^{\infty} \cos^{2n}\theta,\,\,y=\sum_{n=0}^{\infty} \tan^{2n}\psi,\\~~z=\sum_{n=0}^{\infty} \sin^{2n}\theta \tan^{2n}\psi ~~~\left(0<\theta,\psi<\frac{\pi}{4}\right),\,\,$ 

then show that $\,xyz=xy+yz-z.$

Sol. $\,x=\sum_{n=0}^{\infty} \cos^{2n}\theta\\~~~=1+\cos^2\theta+\cos^4\theta+\cdots\\~~~=\frac{1}{1-\cos^2\theta}\\~~~=\frac{1}{\sin^2\theta}\rightarrow(1)$

Again, $y=\sum_{n=0}^{\infty} \tan^{2n}\psi\\~~=1+\tan^2\psi+\tan^4\psi+\cdots\\~~=\frac{1}{1-\tan^2\psi}\rightarrow(2)$

Finally, $\,z=\sum_{n=0}^{\infty} \sin^{2n}\theta \tan^{2n}\psi \\~~=1+\sin^2\theta\tan^2\psi+\sin^4\theta\tan^4\psi+\cdots\\~~=\frac{1}{1-\sin^2\theta\tan^2\psi}\rightarrow(3)$

Now, from $\,(1),\,$ we get, $\,\sin^2\theta=\frac 1x.$

Again, from $\,(2),\,$ we get, $\,1-\tan^2\psi=\frac 1y \\ \Rightarrow \tan^2\psi=1-\frac 1y =\frac{y-1}{y}$

Now, from $\,(3),\,$ we get ,

$\,z=\frac{1}{1-\frac 1x \times \frac{y-1}{y}}\\ \Rightarrow z=\frac{xy}{xy-(y-1)} \\ \Rightarrow xyz-z(y-1)=xy \\ \Rightarrow xyz=xy+yz-z\,\,\text{(showed)}$