Question

Find whether xn + yn is divisible by x-y (y=0) or not
​

Answer 1

Step-by-step explanation:

The statement to be proved is:

P(n):x

n

−y

n

is divisble by x+y where n is even.

Step 1: Verify that the statement is true for the smallest value of n, here, n=2

P(2):x

2

−y

2

is divisible by x+y

P(2):(x+y)(x−y) is divisible by x+y, which is true.

Therefore P(2) is true.

Step 2: Assume that the statement is true for k

Let us assume that P(k):x

k

−y

k

is divisible by x+y where k is even.

Step 3: Verify that the statement is true for the next possible integer, here for n=k+2

x

k+2

−y

k+2

=x

k+2

−x

2

y

k

+x

2

y

k

−y

k+2

=x

2

(x

k

−y

k

)+y

k

(x

2

−y

2

)

Since (x

k

−y

k

) and (x

2

−y

2

) are both divisible by (x+y), the complete equality is divisible by x+y

Therefore,

P(k+2):x

k+2

−y

k+2

is divisible by x+y where k+2 is even.

Therefore by principle of mathematical induction, P(n) is true.

Answer 2

Answer:Let P(n) : xn – yn is divisible by x – y, where x and y are any integers with x≠y.

Now, P(l): x1 -y1 = x-y, which is divisible by (x-y)

Hence, P(l) is true.

Let us assume that, P(n) is true for some natural number n = k.

P(k): xk -yk is divisible by (x – y)

or   xk-yk = m(x-y),m ∈ N …(i)

Now, we have to prove that P(k + 1) is true.

P(k+l):xk+l-yk+l

= xk-x-xk-y + xk-y-yky

= xk(x-y) +y(xk-yk)

= xk(x – y) + ym(x – y)  (using (i))

= (x -y) [xk+ym], which is divisible by (x-y)

Hence, P(k + 1) is true whenever P(k) is true.

So, by the principle of mathematical induction P(n) is true for any natural number n.

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