please solve ........ (mathematic induction class 11th )
Attachments:
Answers
Answered by
1
Hope this will help you !!
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.
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.
Anu726:
Thank you
Similar questions