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Given x so that \(x+x^2+x^3+x^4 = 5\), what is the value of \(\frac{1-x^5}{1-x}\)?

 Oct 10, 2019

Best Answer 

 #2
avatar+30647 
+4

Given

  \(x+x^2+x^3+x^4=5\text{ ...(1)}\\\text{ add 1 to both sides}\\ 1+x+x^2+x^3+x^4=6\text{ ...(2)}\\ \text{ multiply both sides of (2) by x}\\ x+x^2+x^3+x^4+x^5=6x\text{ ...(3)}\\ \text{subtract (3) from (2)}\\1-x^5=6(1-x)\\ \text{Hence : }\frac{1-x^5}{1-x}=6\)

 Oct 10, 2019
 #1
avatar+2856 
+1

Here is my attempt I dont have time I have somwhat a hint
 

Undistribute x + x^2 + x^3 + x^4 = 5

 

x( 1 + x + x^2 + x^3) = 5

 

x( 1 + x( 1 + x + x^2) = 5

 

x( 1 + x( 1 + x(1 + x))) = 5

 

Now I will let anybody else work off from here.

 

 

I think you just keep on dividing by x and subtracting 1 from both sides.

 

Then simplify I am not sure.

 

Good luck everybody!

 Oct 10, 2019
edited by CalculatorUser  Oct 10, 2019
 #2
avatar+30647 
+4
Best Answer

Given

  \(x+x^2+x^3+x^4=5\text{ ...(1)}\\\text{ add 1 to both sides}\\ 1+x+x^2+x^3+x^4=6\text{ ...(2)}\\ \text{ multiply both sides of (2) by x}\\ x+x^2+x^3+x^4+x^5=6x\text{ ...(3)}\\ \text{subtract (3) from (2)}\\1-x^5=6(1-x)\\ \text{Hence : }\frac{1-x^5}{1-x}=6\)

Alan Oct 10, 2019
 #4
avatar+2856 
+1

Thanks, Alan! Another one to study for me. laugh

CalculatorUser  Oct 11, 2019
 #3
avatar+1200 
+1

Thanks guys!

 Oct 10, 2019

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