Question

Find the remainder

when x101 –1 is divided by x -1​

Answer 1

Step-by-step explanation:

What will be the remainder of x101−1 divided by x−1 ?

S[n] = a[x^n -1]/[x-1] is the sum of n terms of a GP , common ratio x

S[100] = 1[x^(101) - 1 ]/[x-1] = 1 + x + x^2 + + + + x^(100) + 0

The quotient ; 1 + x+x^2 + + + x^(100) is a whole number , if x is a whole number , and R =0 .

If the common ratio x is not a whole number then that is a different situation .

There is a problem identifying what the question is

Eg. 1 +5/2 +(5/2) , x=5/2 , n=3 ,a=1

x^n - 1 = [5/2]^3 -1 = [125–8]/8= 117/8

[x^n - 1]/[x-1] = [117/8]/[5/2–1] = [117/8]/[3/2]

What question should be answered?

A;

Is 117/8 to be divided by 3/2 by normal division rules , and get S[3] = 39/4

B;

Divide 117/8 by 3/2 and get some kind of remainder from

[117/8][2/3] ; 234 = 18 mod 24 , or

39/4 ; 39 =3 mod 4

C

What is the remainder when n lots of 3/2 are divided out of 39/4 =9 (3/4]

nq+r=p

n[3/2] +r = 9 (3/4)

6[3/2]+r = 9 (3/4)

r = 9 (3/4) - 9 =3/4

That is 9 (3/4) = (3/4) mod [3/2] ,

That is ; If such an operation , [mod (fraction)] , exists or is tenable

So essentially [x^n - 1]/[x-1] is not a question of remainders or modulo arithmetic , but one of finding the sum of n terms of a GP. Then the question of whether x is a whole number or not does not arise as of being important. Then the quotient 1+x+x^2 + +x^n ,which is itself the sum no more no less no remainder , is just some number whole or not .

hope it will help you