1
Question
xn−yn, n∈N is always divisible by x−y.
Open in App
Solution
The correct option is A True
We can prove this by principle of mathematical induction.Let P(n) be the statement,P(n): xn−yn is divisible by x−y.For n=1,P(1)=x − y, which is obviously divisible by ( x−y).So, P(1) is true.Let us assume that P(k) is true,i.e., xk−yk is divisible by x−y.i.e., xk−yk=t(x−y) for some t∈N. ...(1)We need show that P(k+1) is also true.Now, xk+1−yk+1 =xk+1−x.yk+x.yk−yk+1 =x(xk−yk)+yk(x−y) =x.t(x−y)+yk(x−y) =(x−y)(xt+yk)i.e., xk+1−yk+1=(x−y)(xt+yk)⇒ xk+1−yk+1 is divisible by( x−y).Thus, P(k+1) is true whenever P(k) is true.Hence, by the principle of mathematical induction,P(n) is true for every natural number.
We can prove this by principle of mathematical induction.Let P(n) be the statement,P(n): xn−yn is divisible by x−y.For n=1,P(1)=x − y, which is obviously divisible by ( x−y).So, P(1) is true.Let us assume that P(k) is true,i.e., xk−yk is divisible by x−y.i.e., xk−yk=t(x−y) for some t∈N. ...(1)We need show that P(k+1) is also true.Now, xk+1−yk+1 =xk+1−x.yk+x.yk−yk+1 =x(xk−yk)+yk(x−y) =x.t(x−y)+yk(x−y) =(x−y)(xt+yk)i.e., xk+1−yk+1=(x−y)(xt+yk)⇒ xk+1−yk+1 is divisible by( x−y).Thus, P(k+1) is true whenever P(k) is true.Hence, by the principle of mathematical induction,P(n) is true for every natural number.