Question

A.Find a formula for

1/1×2+1/2×3+...+1/n(n+1)
by examining the values of this expression for small values of n.
b.Prove the formula you conjectured in part (a)

Answer 1

a)
`(n)/(n+1)`

b) Proof in explanation.

Explanation:

a)

`(1)/(1 \cdot 2)+(1)/(2 \cdot 3)+(1)/(3 \cdot 4)+\cdots+(1)/(n(n+1))`.

So let's look at the last term for a minute:

`(1)/(n(n+1))`

Let's use partial fractions to see if we can find a way to write this so it is more useful to us.

`(1)/(n(n+1))=(A)/(n)+(B)/(n+1)`

Multiply both sides by
`n(n+1)`:

`1=A(n+1)+Bn`

Distribute:

`1=An+A+Bn`

Reorder:

`1=An+Bn+A`

Factor:

`1=n(A+B)+A`

This implies
`A=1` and
`A+B=0` which further implies that
`B=-1`.

This means we are saying that:

`(1)/(n(n+1))` can be written as
`(1)/(n)+(-1)/(n+1)`

We can check by combing the fractions:

`(n+1)/(n(n+1))+(-n)/(n(n+1))`

`(n+1-n)/(n(n+1))`

`(1)/(n(n+1))`

So it does check out.

So let's rewrite our whole expression given to us using this:

`((1)/(1)+(-1)/(2))+((1)/(2)+(-1)/(3))+((1)/(3)+(-1)/(4))+\cdots +((1)/(n)+(-1)/(n+1))`

We should see that all the terms in between the first and last are being zeroed out.

That is, this sum is equal to:

`(1)/(1)+(-1)/(n+1)`

Multiply first fraction by (n+1)/(n+1) so we can combine the fractions:

`(n+1)/(n+1)+(-1)/(n+1)`

Combine fractions:

`(n)/(n+1)`

b)

Proof:

Let's see what happens when n=1.

Original expression gives us
`(1)/(1 \cdot 2)=(1)/(2)`.

The expression we came up with gives us
`(1)/(1+1)=(1)/(2)`.

So it is true for the base case.

Let's assume our expression and the expression given is true for some integer k greater than 1.

We want to now show it is true for integer k+1.

So under our assumption we have:

`(1)/(1\cdot 2)+(1)/(2\cdot 3)+\cdots (1)/(k(k+1))=(k)/(k+1)`

So let's add the (k+1)th term of the given series on both sides:

`(1)/(1\cdot 2)+(1)/(2\cdot 3)+\cdots (1)/(k(k+1))+(1)/((k+1)(k+2))=(k)/(k+1)+(1)/((k+1)(k+2))`

(Now we are just playing with right hand side to see if we can put it in the form our solution which be if we can
`(k+1)/(k+2)`.)

I'm going to find a common denominator which will be (k+1)(k+2):

`(k)/(k+1) \cdot (k+2)/(k+2)+(1)/((k+1)(k+2))`

Combine the fractions:

`(k(k+2)+1)/((k+1)(k+2))`

Distribute:

`(k^2+2k+1)/((k+1)(k+2))`

Factor the numerator:

`((k+1)^2)/((k+1)(k+2))`

Cancel a common factor of
`(k+1)`

`(k+1)/(k+2)`

We have proven the given expression and our formula for the sum are equal for all natural numbers,n.