The sum $\frac{1}{1!2004!} + \frac{1}{3!2002!} + \frac{1}{5!2000!} + … + \frac{1}{2003!2!} = \frac{(2^a – 1)}{b!}$. If $a,b$ are positive integers, find $(b-a)$ .
What I Tried :- I saw this Problem in Art of Problem Solving similar to mine :-
If $\frac{1}{0!10!} + \frac{1}{1!9!} + \frac{1}{2!8!} + \frac{1}{3!7!} + \frac{1}{4!6!} + \frac{1}{5!5!}$ is written as a common fraction reduced to lowest terms, the result is $\frac{m}{n}$. Compute the sum of the prime divisors of $m$ plus the sum of the prime divisors of $n$.
I saw the solution written like this here and tried to understand it but couldn't .
The first thing is to notice how much like binomial coefficients these numbers look like. Thus we can write the expression $\frac{1}{0!10!} + \frac{1}{1!9!} + \frac{1}{2!8!} + \frac{1}{3!7!} + \frac{1}{4!6!} + \frac{1}{5!5!}$ as:
$$\frac{1}{10!}\left(\binom{10}{0}+\binom{10}{9}+\binom{10}{8}+\binom{10}{7}+\binom{10}{6}+\binom{10}{5}\right).$$
Call the sum inside the parentheses $S$. Now notice that $$2S=\left(\binom{10}{0}+\binom{10}{1}+\cdots+\binom{10}{10}\right)+\binom{10}{5}.$$ Using a well known combinatorial identity we get $2S=2^{10}+\binom{10}{5}=1276\Rightarrow S=638.$ Thus we get $\frac{638}{10!}=\frac{2\cdot 11\cdot 29}{10!}.$ And our answer is $11+29+2+3+5+7=057.$
Can someone explain why this is happening? Also can someone say how can you understand that you have to factor $\frac{1}{10!}$ out of it? $$\frac{1}{10!}\left(\binom{10}{0}+\binom{10}{9}+\binom{10}{8}+\binom{10}{7}+\binom{10}{6}+\binom{10}{5}\right).$$
If someone can explain this, probably I will be able to solve my problem?
Best Answer
Here is a solution, nicely detailed.
$$\sum_{i=1}^{2003}\frac{1}{i!(2005-i)!}=\frac{1}{2005!}\cdot \sum_{i=1}^{2003}\frac{2005!}{i!(2005-i)!}=\frac{1}{2005!}\cdot \sum_{i=1}^{2003}\binom{2005}{i}=\frac{1}{2005!}\cdot\big(2^{2005}-2007\big)=\frac{2^2005-2007}{2005!}$$
Why does this happen? From Newton's formula, we have
$$(x+y)^n=\sum_{i=0}^{n}\bigg(\binom{n}{i}\cdot x^i\cdot y^{n-i}\bigg)$$
so for $x=y=1$ we have
$$2^n=\sum_{i=0}^{n}\binom{n}{i}$$
thus giving us
$$\sum_{i=1}^{2003}\binom{2005}{i}=2^{2005}-\binom{2005}{0}-\binom{2005}{2004}-\binom{2005}{2005}=2^{2005}-2007$$
So in fact, your statement is wrong. For it to have been right, it should have said:
$$\sum_{i=0}^{2004}\frac{1}{i!(2005-i)!}=\frac{2^a-1}{b}$$ or $$\sum_{i=1}^{2005}\frac{1}{i!(2005-i)!}=\frac{2^a-1}{b}$$.
In both cases, using the exact same approach, our sum is equal to $$\frac{2^{2005}-1}{2005!}$$ so $a-b=0$