1. ## Binomial expansion

If $\displaystyle S_{n}=\frac{^{n}c_{1}}{1}-\frac{^{n}c_{2}}{2}+\frac{^{n}c_{3}}{3}-\frac{^{n}c_{4}}{4}+..........+\frac{(-1)^{n-1} \ ^{n}c_{n}}{n}$

Show that
$\displaystyle S_{n+1}-S_{n}=\frac{1}{n+1}$

I go by this way>>>>

n----> n+1

$\displaystyle S_{n+1}=\frac{^{n+1}c_{1}}{1}-\frac{^{n+1}c_{2}}{2}+\frac{^{n+1}c_{3}}{3}-\frac{^{n+1}c_{4}}{4}+..........+(-1)^{n-1}\frac{^{n-1}c_{n}}{n}+(-1)^{n}\frac{^{n+1}c_{n+1}}{n+1}$
After that
$\displaystyle S_{n+1}=\frac{^{n}c_{1}+^{n}c_{0}}{1}-\frac{^{n}c_{2}+^{n}c_{1}}{2}+\frac{^{n}c_{3}+^{n} c_{2}}{3}-\frac{^{n}c_{4}+^{n}c_{3}}{4}+...................+ (-1)^{n-1}\frac{^{n-1}c_{n}+^{n-1}c_{n-1}}{1}+(-1)^{n}\frac{^{n}c_{n+1}+^{n}c_{n}}{n+1}$
But when i subtract i could not get like that>>>>>

2. ## Re: Binomial expansion

Your notation is slightly confusing. The title of your post "Binomial Expansion" could suggest two possible ways to interpret your notation. You might be using $\displaystyle {^n}C_r = \dfrac{n!}{(n-r)!r!}$, which is a binomial coefficient, or you may be asking that we help you expand the following using Newton's Generalized Binomial Formula:

$\displaystyle {^nc} = \underbrace{c^{c^{.^{.^{.^c}}}}}_{n\text{ times}}$

I believe you mean the former, so I will change the notation slightly:

$\displaystyle S_n = \frac{{_n}C_1}{1} - \frac{{_n}C_2}{2} + \frac{{_n}C_3}{3} - \frac{{_n}C_4}{4} + \cdots + \frac{(-1)^{n-1}{_n}C_n}{n}$

where currently, we assume $\displaystyle {_n}C_r$ is only defined when $\displaystyle n,r$ are nonnegative integers. Then $\displaystyle {_n}C_r = \begin{cases}\dfrac{n!}{r!(n-r)!} & \text{if }n\ge r \\ 0 & \text{otherwise}\end{cases}$

So, $\displaystyle S_{n+1} = \sum_{r=1}^{n+1} (-1)^{r-1}\dfrac{{_{n+1}}C_r}{r}$ (as you wrote in your post).

However, let's manipulate this a little before we subtract $\displaystyle S_n$. Using Pascal's Rule, which is $\displaystyle {_{n+1}}C_r = {_n}C_{r-1}+{_n}C_r$, we get:

\displaystyle \begin{align*}S_{n+1} & = \sum_{r=1}^{n+1} (-1)^{r-1}\dfrac{{_{n+1}}C_r}{r} \\ & = \sum_{r=1}^{n+1} (-1)^{r-1} \dfrac{{_n}C_{r-1} + {_n}C_r}{r} \\ & = \left( \sum_{r=1}^{n+1} (-1)^{r-1} \dfrac{{_n}C_{r-1}}{r} \right) + \left( \sum_{r=1}^{n+1} (-1)^{r-1} \dfrac{{_n}C_r}{r} \right) \\ & = \left( \sum_{r=1}^{n+1} (-1)^{r-1} \dfrac{{_n}C_{r-1}}{r} \right) + \left( \sum_{r=1}^n (-1)^{r-1} \dfrac{{_n}C_r}{r} \right) + (-1)^{n+1-1} \dfrac{{_n}C_{n+1}}{n+1} \\ & = \left( \sum_{r=1}^{n+1} (-1)^{r-1} \dfrac{{_n}C_{r-1}}{r} \right) + \left( \sum_{r=1}^n (-1)^{r-1} \dfrac{{_n}C_r}{r} \right) + (-1)^n \dfrac{0}{n+1} \\ & = \left( \sum_{r=1}^{n+1} (-1)^{r-1} \dfrac{{_n}C_{r-1}}{r} \right) + S_n\end{align*}

Subtracting $\displaystyle S_n$ from both sides gives:

\displaystyle \begin{align*}S_{n+1} - S_n & = \sum_{r=1}^{n+1} (-1)^{r-1} \dfrac{{_n}C_{r-1}}{r} \\ & = \sum_{r=1}^{n+1} (-1)^{r-1} \dfrac{n!}{(n+1-r)!r!} \\ & = n!\sum_{r=1}^{n+1} (-1)^{r-1} \dfrac{1}{(n+1-r)!r!} \\ & = n!\left(\dfrac{1}{n!1!} - \dfrac{1}{(n-1)!2!} + \cdots + (-1)^{n-2}\dfrac{1}{2!(n-1)!} + (-1)^{n-1}\dfrac{1}{1!n!} + (-1)^n\dfrac{1}{0!(n+1)!} \right) \end{align*}

Let's look at this for small values of $\displaystyle n$.

(n=1):
$\displaystyle S_2-S_1 = 1!\left(\dfrac{1}{1!1!} - \dfrac{1}{0!2!} \right) = \dfrac{1}{2} = \dfrac{1}{1+1}$

(n=2):
$\displaystyle S_3 - S_2 = 2!\left(\dfrac{1}{2!1!} - \dfrac{1}{1!2!} + \dfrac{1}{0!3!} \right) = 2!\left(0 + \dfrac{1}{3!}\right) = \dfrac{1}{3} = \dfrac{1}{2+1}$

(n=3):
\displaystyle \begin{align*}S_4 - S_3 & = 3!\left(\dfrac{1}{3!1!} - \dfrac{1}{2!2!} + \dfrac{1}{1!3!} - \dfrac{1}{0!4!} \right) \\ & = 3!\left( 2\dfrac{1}{3!}-\dfrac{1}{2!2!}-\dfrac{1}{4!} \right) \\ & = 2-\dfrac{3}{2}-\dfrac{1}{4} = \dfrac{1}{4} = \dfrac{1}{3+1}\end{align*}

(n=4):
\displaystyle \begin{align*}S_5 - S_4 & = 4!\left( \dfrac{1}{4!1!}-\dfrac{1}{3!2!}+\dfrac{1}{2!3!} - \dfrac{1}{1!4!} + \dfrac{1}{0!5!} \right) \\ & = 4!\left(0+\dfrac{1}{5!}\right) = \dfrac{1}{5} = \dfrac{1}{4+1}\end{align*}

The pattern is that when $\displaystyle n$ is even, the first $\displaystyle n$ terms cancel each other out. When $\displaystyle n$ is odd, the first $\displaystyle n$ terms should add to $\displaystyle \dfrac{2}{n+1}$.

To see that this is true, suppose $\displaystyle n$ is even. Then we have:

\displaystyle \begin{align*}\sum_{r=1}^{n+1} (-1)^{r-1} \dfrac{n!}{(n+1-r)!r!} & = \sum_{r=1}^{n/2}(-1)^{r-1}\dfrac{n!}{(n+1-r)!r!} + \sum_{r=n/2+1}^n (-1)^{r-1}\dfrac{n!}{(n+1-r)!r!} + (-1)^n\dfrac{n!}{0!(n+1)!}\\ & = \left(\sum_{r=1}^{n/2} (-1)^{r-1}\dfrac{n!}{(n+1-r)!r!} + \sum_{r=1}^{n/2}(-1)^{n-r}\dfrac{n!}{r!(n+1-r)!}\right) + \dfrac{1}{n+1}\end{align*}

When $\displaystyle r$ is odd, then $\displaystyle r-1$ is even and $\displaystyle n-r$ is odd, so $\displaystyle (-1)^{r-1}\dfrac{n!}{(n+1-r)!r!} + (-1)^{n-r}\dfrac{n!}{r!(n+1-r)!} = \dfrac{n!}{(n+1-r)!r!} - \dfrac{n!}{r!(n+1-r)!} = 0$. When $\displaystyle r$ is even, then $\displaystyle r-1$ is odd and $\displaystyle n-r$ is even, so $\displaystyle (-1)^{r-1}\dfrac{n!}{(n+1-r)!r!} + (-1)^{n-r}\dfrac{n!}{r!(n+1-r)!} = -\dfrac{n!}{(n+1-r)!r!}+\dfrac{n!}{r!(n+1-r)!} = 0$. Hence, the only term that remains is the $\displaystyle \dfrac{1}{n+1}$.

See if you can figure out how to show that when $\displaystyle n$ is odd, you get $\displaystyle \dfrac{2}{n+1}$.