use principle of M.I to show that 1÷1.2+1÷2.3+1÷3.4...+ 1÷(n+1)(n+2)=n+1÷n+2?

To prove this using the method of mathematical induction, we first need to show that the statement is true for the base case, which is n = 1.

When n = 1:
1÷1.2 = 1 = 1+1÷2 = 1+1÷2

Now, assume that the statement is true for some arbitrary positive integer k, i.e.,

1÷1.2 + 1÷2.3 + ... + 1÷(k+1)(k+2) = k+1÷k+2

We need to show that the statement is also true for k+1, i.e.,

1÷1.2 + 1÷2.3 + ... + 1÷(k+1)(k+2) + 1÷(k+2)(k+3) = k+2÷k+3

Starting with the left-hand side of the equation:

1÷1.2 + 1÷2.3 + ... + 1÷(k+1)(k+2) + 1÷(k+2)(k+3) = (k+1)÷(k+2) + 1÷(k+2)(k+3)

= (k+1)(k+3)÷(k+2)(k+3) + 1÷(k+2)(k+3)

= (k^2 + 4k + 3)÷(k^2 + 5k + 6)

= (k+3)(k+1)÷(k+3)(k+2)

= k+2÷k+3

Therefore, by the principle of mathematical induction, the statement 1÷1.2 + 1÷2.3 + ... + 1÷(n+1)(n+2) = n+1÷n+2 is true for all positive integers n.