>
#mathematics
,
#principles-of-mathematics
,
#mathematical-induction
>
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?

Feb 24

Mth 101

Sort By

ChatGPT

Feb 24

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.

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.