Statement
xn−yn=(x−y)(xn−1+xn−2y+⋯+xyn−2+yn−1)
Proof
Suppose n=1:
x−y=(x−y)∗1=x−y
Suppose the statement is true for all k < n:
(xk−yk)(x+y)=xk+1+xky−xyk−yk+1
xk+1−yk+1=(xk−yk)(x+y)−xky+xyk
xk+1−yk+1=(x−y)(xk−1+xk−2y⋯+xyk−2+yk−1)(x+y)−xy(xk−1+yk−1)
xk+1−yk+1=(x−y)(xk−1+xk−2y⋯+xyk−2+yk−1)(x+y)−xy(xk−1−yk−1)
xk+1−yk+1=(x−y)(xk−1+xk−2y⋯+xyk−2+yk−1)(x+y)−xy(x−y)(xk−2+xk−3y+⋯+xyk−3+yk−2)
xk+1−yk+1=(x−y)(xk−1+xk−2y⋯+xyk−2+yk−1)(x+y)−(x−y)(xk−1+xk−2y2+⋯+x2yk−2+yk−1)
xk+1−yk+1=(x−y)(xk+xk−1y+⋯+x2yk−2+xyk−1+xk−1y+xk−2y2+⋯+xyk−1+yk−xk−1y−xk−2y2−⋯−x2yk−2−xyk−1)
xk+1−yk+1=(x−y)(xk+xk−1y+⋯+x2yk−2+xyk−1+yk)■
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