GO Classes CS/DA 2025 | Weekly Quiz 1 | Fundamental Course | Question: 3

Answer 1

Comments

If you solve the gp on RHS, you will get a/1-r as 1/1-x and you have another (1-x) on LHS so bringing that one to RHS will give you 1/(1-x^2)

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( 1-x)S = 1+x^2+x^3+x^4+x^5+…………
 => S = 1/(1-x) * 1+x^2+x^3+x^4+…..

how did you get S = 1/(1-x)^2.

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( 1-x)S = 1+x^2+x^3+x^4+x^5+…………

Bro, What will be the summation of  $1$ $+$ $x^2$ $+$ $x^3$ +$x^4$ $+$ ………….?

 

Answer 2

let ${S=1+2x+3x}$$^{2}$${+4x}$$^{3}$${+…..}$

given ${|x|<1} $    i.e ${-1<x<1}$ hence ${S }$ is not infinite because if we take power of number (-1,1) so it is smaller than its original that is power do not increases this number but if x >1 or x<-1 then x to power will become grater and since this series is till infinite then there sum is obviously infinite so no reason to find such summation but here we know that it is not infinite rather than that it is .

 

now  

${S=1+2x+3x}$$^{2}$${+4x}$$^{3}$${+…..}$

${xS=x+2x}$$^{2}$${+3x}$$^{3}$${+4x}$$^{4}$${+…..}$

so

${S-xS=1+2x+3x}$$^{2}$${+4x}$$^{3}$${+…..- (x+2x}$$^{2}$${+3x}$$^{3}$${+4x}$$^{4}$${+…..)}$

${S-xS=1+x+x}$$^{2}$${+x}$$^{3}$${+….}$

let    ${T=S-xS=S(1-x)=1+x+x}$$^{2}$${+x}$$^{3}$${+….}$

now ${xT=x+x}$$^{2}$${+x}$$^{3}$${+….}$

${T-xT =1+x+x}$$^{2}$${+x}$$^{3}$${+….-(x+x}$$^{2}$${+x}$$^{3}$${+….)}$

${T(1-x)=1}$

since ${T=S(1-x)}$

hence ${S(1-x)(1-x)=1}$

${S(1-x)}$$^{2}$${=1}$

hence ${S=1/(1-x)}$$^{2}$ where ${|x|<1}$
${Ans .  A}$