Question

plzz solve this fast....prove that (x ^n-1)is divisible by (x-1)​ with the help of mathematical induction.​

Answer 1

Answer:

Step-by-step explanation:

Pu.nnet

Answer 2

So first you can't assume that the left hand side is divisible by x−1x−1 but for the right hand side we have that x−1x−1 divides x−1x−1 and by the induction hypothesis we have that x−1x−1 divides xn−1−1xn−1−1 so what can you conclude about the left hand side.

well, if we assume xn−1−1xn−1−1 is divisible by x−1x−1, then is x(xn−1−1)x(xn−1−1) also divisible?

You seem to be trying to prove exactly the opposite of what you should be proving. In effect you’re assuming that xn−1xn−1 is divisible by x−1x−1 and showing that xn−1−1xn−1−1 is also divisible by x−1x−1. You should be going in the other direction: for the induction step you want to show that if xn−1xn−1 is divisible by x−1x−1, then xn+1−1xn+1−1 is also divisible by x−1x−1. With very minor modification you already have the necessary algebra:

xn+1−1=x(xn−1)+(x−1).

xn+1−1=x(xn−1)+(x−1).

How can you use this to show that if xn−1xn−1 is divisible by x−1x−1, then so is xn+1−1xn+1−1?

let P(n)P(n) be the assertion that xn−1xn−1 is divisible by x−1x−1.

we know that P(0)P(0) is true. suppose that for some nn we have P(n)P(n). then there is a number kk such that:

xn−1=k(x−1)

xn−1=k(x−1)

now add xn+1−xn=xn(x−1)xn+1−xn=xn(x−1) to both sides of the equation. this will give you the inductive step from P(n)P(n) to P(n+1)P(n+1).