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.
Best Answer
It is convenient to use the coefficient of operator $[x^r]$ to denote the coefficient of a series.
In (1) we select the coefficient of $x^r$.
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*}