Question

A.Find a formula for

Answer 1

Answer:

a) $\frac{n}{n+1}$

b) Proof in explanation.

Step-by-step explanation:

a)

$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\cdots+\frac{1}{n(n+1)}$.

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

$\frac{1}{n(n+1)}$

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

$\frac{1}{n(n+1)}=\frac{A}{n}+\frac{B}{n+1}$

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:

$\frac{1}{n(n+1)}$ can be written as $\frac{1}{n}+\frac{-1}{n+1}$

We can check by combing the fractions:

$\frac{n+1}{n(n+1)}+\frac{-n}{n(n+1)}$

$\frac{n+1-n}{n(n+1)}$

$\frac{1}{n(n+1)}$

So it does check out.

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

$(\frac{1}{1}+\frac{-1}{2})+(\frac{1}{2}+\frac{-1}{3})+(\frac{1}{3}+\frac{-1}{4})+\cdots +(\frac{1}{n}+\frac{-1}{n+1})$

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

That is, this sum is equal to:

$\frac{1}{1}+\frac{-1}{n+1}$

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

$\frac{n+1}{n+1}+\frac{-1}{n+1}$

Combine fractions:

$\frac{n}{n+1}$

b)

Proof:

Let's see what happens when n=1.

Original expression gives us $\frac{1}{1 \cdot 2}=\frac{1}{2}$.

The expression we came up with gives us $\frac{1}{1+1}=\frac{1}{2}$.

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:

$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots \frac{1}{k(k+1)}=\frac{k}{k+1}$

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

$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots \frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}$

(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 $\frac{k+1}{k+2}$.)

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

$\frac{k}{k+1} \cdot \frac{k+2}{k+2}+\frac{1}{(k+1)(k+2)}$

Combine the fractions:

$\frac{k(k+2)+1}{(k+1)(k+2)}$

Distribute:

$\frac{k^2+2k+1}{(k+1)(k+2)}$

Factor the numerator:

$\frac{(k+1)^2}{(k+1)(k+2)}$

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

$\frac{k+1}{k+2}$

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