Find the quotient (a) x^n-1 ÷ x-1 for n = 2, 3, 4, and 8 (b) What patterns do you notice? (c) Use your work in part (a) to write an expression equivalent to x^n-1 ÷ x-1 for any integer n > 1.

(b) when we divide ([tex]x^{n} -1[/tex]) with (x-+1) then the quotient will be a polynomial of x with (n-1) degree and all the coefficients are 1.

(c) [tex][1+x^{2} +x^{3}+x^{4}+x^{5}+.....+x^{n-2} +x^{n-1} ][/tex]

Step-by-step explanation:

(a) We have, [tex]\frac{x^{2} -1}{x-1} =\frac{(x-1)(x+1)}{x-1} =1+x[/tex] (Answer)

and, [tex]\frac{x^{3} -1}{x-1} =\frac{(x-1)(x^{2}+x+1) }{x-1}=1+x+x^{2}[/tex] (Answer)

and, [tex]\frac{x^{4}-1 }{x-1} =\frac{(x-1)(x+1)(x^{2}+1) }{x-1}=(x+1)(x^{2} +1) =x^{3}+x^{2} +x+1[/tex] (Answer)

and, [tex]\frac{x^{8}-1 }{x-1} = \frac{(x-1)(x+1)(x^{2}+1)(x^{4} +1) }{x-1} =(x+1)(x^{2}+1)(x^{4} +1) =x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+1[/tex] (Answer)

(b) From the above four quotients it is clear that when we divide ([tex]x^{n} -1[/tex]) with (x-+1) then the quotient will be a polynomial of x with (n-1) degree and all the coefficients are 1. (Answer)

(c) Hence, from the above pattern of the quotients we can write the expression which is equivalent to [tex]\frac{x^{n}-1 }{x-1}[/tex] will be

[tex][1+x^{2} +x^{3}+x^{4}+x^{5}+.....+x^{n-2} +x^{n-1} ][/tex] (Answer)