- #1

opus

Gold Member

- 717

- 131

Express the sum ##\frac{1}{2⋅3}+\frac{1}{3⋅4}+\frac{1}{4⋅5}+...+\frac{1}{2019⋅2020}## as a fraction of whole numbers in lowest terms.

It goes on to state that each term in the sum is of the form ##\frac{1}{k\left(k+1\right)}## which is obvious.

The partial fraction decomposition of this is ##\frac{1}{k\left(k+1\right)}=\frac{1}{k}-\frac{1}{k+1}## Again, no problems here.

Now it follows that by using this decomposition for each term, we get:

##\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+...+\left(\frac{1}{2019}-\frac{1}{2020}\right)##

After removing the parenthesis, the only terms that are left, after adding them all up, is the first and last term.

My question is, in being presented with this problem, what "flags" if you will, should we be looking that tells us to use partial fraction decomposition? In trying this problem myself, I got to just the basic setup of ##\frac{1}{k\left(k+1\right)}##, but from here, I would never guess that it needed to be decomposed.