why multiplying the two series together give this as the coefficient of x^r?

Question

If $c_1,c_2,.....c_n$ are the coefficients in the expansion of $(1+x)^n$, when n is a positive integer, prove that

$c_0-c_1+c_2-c_3+........+(-1)^rc_r=(-1)^r\frac{n!-1}{r!(n-r-1)!}$

Here is the solution (it's an image).

I am having trouble understanding why multiplying the two series together give this as the coefficient of $x^r$?

Can anyone explain how we get this?

Any help would be appreciated.

Answer 1

Here one has to prove that $$S=\sum_{k=0}^{r} (-1)^k {n \choose k}= (-1)^r {n-1 \choose r}$$, there are several methods to prove this sum. But let us take the route mentioned by OP. $$(1+x)^n=C_0+C_1 x+ C_2 x^2 +C_3 x^3+ ......+C_{r-3} x^{r-3}+ C_{r-2} x^{r-2}+ C_{r-1} x^{r-1}+ C_r x^r+....+C_n x^n....(1).$$ Here $C_k={n \choose k}.$ The infinite G.P is written as $$(1+x)^{-1}=1-x+x^2-x^3+x^4-x^5+....+(-1)^{r-2} x^{r-2}+ (-1)^{r-1} x^{r-1}+(-1)^r x^{r}+....+...~~~(2)$$ Now we multiply one term from (1) and one from (2) so that product of the two terms has $x^r$ in it, so $$(1+x)^{n} (1+x)^{-1}=[C_0 (-1)^r ~ x^r + C_1 x (-1)^{r-1} x^{r-1}+ C_2 (-1)^{r-2} x^{r-2}+....+C_r x^r]+......\text{many other terms}$$ $$\implies (1+x)^{n-1}=(-1)^r[C_0-C_1 +C_2 -C_3+.....+(-1)^r C_r] x^r+---\text{many other terms}.$$ Thus, $$(-1)^r[C_0-C_1 + C_2-C_3+.....+(-1)^r C_r]= (-1)^t~S= \text{co-efficient of}~ x^r ~ \text{in}~ (1+x)^{n-1}= {n-1 \choose r}.$$ Finally, $$S=(-1)^r ~ {n-1 \choose r}$$

Answer 2

It is convenient to use the coefficient of operator $[x^r]$ to denote the coefficient of a series.

We obtain the coefficient of $x^r$ of $(1+x)^{-1}(1+x)^n$ in two different ways. At first we multiply out the terms. \begin{align*} \color{blue}{[x^r](1+x)^{-1}(1+x)^n}&=[x^r](1+x)^{n-1}\\ &\,\,\color{blue}{=\binom{n-1}{r}}\tag{1} \end{align*}

In (1) we select the coefficient of $x^r$.

Now we expand at first the geometric series $(1+x)^{-1}=1-x+x^2-x^3+\cdots$ and obtain \begin{align*} \color{blue}{[x^r](1+x)^{-1}(1+x)^n}&=[x^r]\sum_{j=0}^\infty (-1)^jx^j(1+x)^n\tag{2}\\ &=\sum_{j=0}^r(-1)^j[x^{r-j}](1+x)^n\tag{3}\\ &=\sum_{j=0}^r(-1)^{r-j}[x^j](1+x)^n\tag{4}\\ &\,\,\color{blue}{=(-1)^r\sum_{j=0}^{r}(-1)^j\binom{n}{j}}\tag{5} \end{align*}

Comparing (1) and (5) shows the claim is valid.

Comment:

• In (2) we expand the geometric series $\frac{1}{1+x}=\sum_{j=0}^\infty (-1)^jx^j$.

• In (3) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$. We set the upper index of the sum to $r$ since other indices do not contribute to $[x^r]$.

• In (4) we change the order of summation $j\to r-j$.

• In (5) we select the coefficient of $x^j$.

Supplement regarding the alternating sum $\color{blue}{(-1)^r\{c_0-c_1+c_2-c_3+........+(-1)^rc_r\}}$:

• Recalling the product of two series $C(x)=\sum_{k=0}^\infty c_kx^k$ with $A(x)=\sum_{j=0}^\infty a_jx^j$ we can interpret the multiplication with $A(x)$ as transformation of the coefficients $c_k$ \begin{align*} C(x)A(x)&=\left(\sum_{k=0}^\infty c_kx^k\right)\left(\sum_{j=0}^\infty a_jx^j\right)\\ &=\sum_{r=0}^\infty\left(\sum_{k=0}^rc_ka_{r-k}\right)x^r\tag{6}\\ \\ \color{blue}{c_r}\quad&\color{blue}{\to\quad\sum_{k=0}^rc_ka_{r-k}}\tag{7} \end{align*}

• A special case of (7) is to obtain the sum of the first $r+1$ coefficients of $C(x)$ by setting $A(x)=\frac{1}{1-x}$. We obtain from (6) \begin{align*} C(x)\frac{1}{1-x}&=C(x)\sum_{j=0}^\infty x^j\\ &=\sum_{r=0}^\infty\left(\sum_{k=0}^rc_k\right)x^r\\ \color{blue}{c_r}\quad&\color{blue}{\to\quad\sum_{k=0}^rc_k=c_0+c_1+c_2+\cdots+c_r} \end{align*}

• Another special case of (7) is to obtain the alternating sum of the first $r+1$ coefficients of $C(x)$ by setting $A(x)=\frac{1}{1+x}$. We obtain from (6) \begin{align*} C(x)\frac{1}{1+x}&=C(x)\sum_{j=0}^\infty (-1)^jx^j\\ &=\sum_{r=0}^\infty\left(\sum_{k=0}^rc_k(-1)^{r-k}\right)x^r\\ \color{blue}{c_r}\quad&\color{blue}{\to\quad(-1)^r\sum_{k=0}^r(-1)^kc_k=(-1)^r\left\{c_0-c_1+c_2-\cdots+(-1)^rc_r\right\}} \end{align*}