## Sunday, May 24, 2015

### Method 2: Find x and y if $\dfrac{1}{1!21!}+\dfrac{1}{3!19!}+\dfrac{1}{5!17!}+\cdots+\dfrac{1}{21!1!}=\dfrac{2^x}{y!}$

If $x,\,y$ are positive integers such that $\dfrac{1}{1!21!}+\dfrac{1}{3!19!}+\dfrac{1}{5!17!}+\dfrac{1}{7!15!}+\dfrac{1}{9!13!}+\dfrac{1}{11!11!}+\dfrac{1}{13!9!}+\dfrac{1}{15!7!}+\dfrac{1}{17!5!}+\dfrac{1}{19!3!}+\dfrac{1}{21!1!}=\dfrac{2^x}{y!}$. Find $x,\,y$.

Method 2:

Method 1 has showed us the given LHS of the equation (after it's multiplied both sides by the quantity $22!$) is equivalent to

[MATH] {22 \choose 1}+{22 \choose 3}+{22 \choose 5}+{22 \choose 7}+{22 \choose 9}+{22 \choose 11}+{22 \choose 13}+{22 \choose 15}+{22 \choose 17}+{22 \choose 19}+{22 \choose 21}=\dfrac{2^x22!}{y!}[/MATH]

This second method doesn't require us to use calculator to look for both $x$ and $y$. All it needs is the binomial theorem, that would be all.

The formal expression of the Binomial Theorem is as follows:

[MATH](a+b)^n = \sum_{k=0}^n{n \choose k}a^{n-k}b^k[/MATH]

What is so significant about this binomial theorem is on what figures we replace $a$ and $by$ with.

If we let $a=b=1$, we have:

[MATH]\begin{align*}(1+1)^n& = \sum_{k=0}^n{n \choose k}1^{n-k}1^k\\&= \sum_{k=0}^n{n \choose k}\\&={n \choose 0}+{n \choose 1}+{n \choose 2}+{n \choose 3}\cdots+{n \choose n-1}+{n \choose n}\end{align*}[/MATH]

On the other hand, if we let $a=1$ and $b=-1$, we obtain:

[MATH]\begin{align*}(1-1)^n& = \sum_{k=0}^n{n \choose k}1^{n-k}(-1)^k\\&= \sum_{k=0}^n(-1)^k{n \choose k}\\&={n \choose 0}-{n \choose 1}+{n \choose 2}-{n \choose 3}\cdots-+{n \choose n-1}+{n \choose n}
\color{red}\text{*(if $n$ is even)}\end{align*}[/MATH]

In our case, we have $n=22$.

Therefore we get:

[MATH](1+1)^{22}=2^{22}={22 \choose 0}+{22 \choose 1}+{n \choose 2}+{n \choose 3}+\cdots+{22 \choose 21}+{22 \choose 22}[/MATH]

[MATH](1-1)^{22}=0={22 \choose 0}-{22 \choose 1}+{22 \choose 2}-{22 \choose 3}\cdots-+{22 \choose 21}+{22 \choose 22}[/MATH]

Equation below subtracted from the above gives:

[MATH]2^{22}=2\left({22 \choose 1}+{22 \choose 3}+{22 \choose 5}+{22 \choose 7}+{22 \choose 9}+{22 \choose 11}+{22 \choose 13}+{22 \choose 15}+{22 \choose 17}+{22 \choose 19}+{22 \choose 21}\right)[/MATH]

[MATH]2^{21}={22 \choose 1}+{22 \choose 3}+{22 \choose 5}+{22 \choose 7}+{22 \choose 9}+{22 \choose 11}+{22 \choose 13}+{22 \choose 15}+{22 \choose 17}+{22 \choose 19}+{22 \choose 21}[/MATH]

Therefore we have:

$2^{21}=\dfrac{2^x22!}{y!}$

$2^{21}y!=2^x22!$

Hence $x=21$ and $y=22$ are the solutions.