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# last question team!

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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

#2
+29243
+4

Given

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

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Oct 10, 2019

#1
+2847
+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
+29243
+4

Given

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

Alan Oct 10, 2019
#4
+2847
+1

Thanks, Alan! Another one to study for me.

CalculatorUser  Oct 11, 2019
#3
+1196
+1

Thanks guys!

Oct 10, 2019