value of $k$ in binomial sum

Question

If $\displaystyle (1-x)^{\frac{1}{2}}=a_{0}+a_{1}x+a_{2}x^2+\cdots \cdots +\infty.$ and $a_{0}+a_{1}+a_{2}+\cdots+a_{10}=\frac{\binom{20}{10}}{k^{10}}.$ Then $k$ is

what i try

$(1-x)^{\frac{1}{2}}=1-\frac{1}{2}x+\frac{1}{2}\bigg(\frac{1}{2}-1\bigg)\frac{x^2}{2!}+\frac{1}{2}\bigg(\frac{1}{2}-1\bigg)\bigg(\frac{1}{2}-2\bigg)\frac{x^3}{3!}+\cdots\cdots$

$\displaystyle a_{0}=1,a_{1}=-\frac{1}{2},a_{2}=-\frac{1}{2^2}\frac{1}{2!},a_{3}=\frac{1\cdot 3}{2^3}\frac{1}{3!}+\cdots$

How do i solve it Help me pleses

Answer 1

We have $$\sum_{n=0}^{1}(-1)^n\binom{\frac 12}{n}=\frac 12=\frac 24=\frac{\binom{2}{1}}{4^1},\quad \sum_{n=0}^{2}(-1)^n\binom{\frac 12}{n}=\frac 38=\frac{6}{16}=\frac{\binom{4}{2}}{4^2}$$ $$\sum_{n=0}^{3}(-1)^n\binom{\frac 12}{n}=\frac{5}{16}=\frac{20}{64}=\frac{\binom{6}{3}}{4^3},\quad \sum_{n=0}^{4}(-1)^n\binom{\frac 12}{n}=\frac{35}{128}=\frac{70}{256}=\frac{\binom{8}{4}}{4^4}$$ So, it seems that the following holds : $$\sum_{n=0}^{r}(-1)^n\binom{\frac 12}{n}=\frac{\binom{2r}{r}}{4^r}\tag1$$

One can prove $(1)$ by induction.

Proof :

$(1)$ holds for $r=0$.

Supposing that $(1)$ holds for $r$ gives

$$\begin{align}\sum_{n=0}^{r+1}(-1)^n\binom{\frac 12}{n}&=\sum_{n=0}^{r}(-1)^n\binom{\frac 12}{n}+(-1)^{r+1}\binom{\frac 12}{r+1} \\\\&=\frac{\binom{2r}{r}}{4^r}+(-1)^{r+1}\cdot\frac{\frac 12(\frac 12-1)\cdots (\frac 12-r)}{(r+1)!} \\\\&=\frac{\binom{2r}{r}}{4^r}+(-1)^{r+1}\cdot\frac{(2r-1)(2r-3)\cdots (2-1)(-1)}{(r+1)!(-2)^{r+1}} \\\\&=\frac{\binom{2r}{r}}{4^r}+(-1)^{r+1}\cdot\frac{-(2r-1)(2r-3)\cdots (2-1)\cdot (2r)(2r-2)\cdots 2}{(r+1)!(-2)^{r+1}\cdot (2r)(2r-2)\cdots 2} \\\\&=\frac{\binom{2r}{r}}{4^r}+(-1)^{r+1}\cdot\frac{-(2r)!}{(r+1)!(-2)^{r+1}\cdot 2^rr!} \\\\&=\frac{\binom{2r}{r}}{4^r}+\frac{-1}{(r+1)\cdot 2^{2r+1}}\cdot\frac{(2r)!}{r!r!} \\\\&=\frac{\binom{2r}{r}}{4^r}\bigg(1-\frac{1}{2(r+1)}\bigg) \\\\&=\frac{1}{4^r}\cdot\frac{(2r)!}{r!r!}\cdot\frac{2r+1}{2r+2}\cdot\frac{2r+2}{2r+2} \\\\&=\frac{\binom{2r+2}{r+1}}{4^{r+1}}\qquad\square\end{align}$$

It follows from $(1)$ that $$a_0+a_1+\cdots +a_{10}=\sum_{n=0}^{10}(-1)^n\binom{\frac 12}{n}=\frac{\binom{20}{10}}{\color{red}4^{10}}$$

Answer 2

The binomial expansion is $$\sqrt{1-x}=\sum_{n=0}^\infty (-1)^n\binom{\frac{1}{2}}{n}x^n$$ and so you need to solve for $k$ $$\sum_{n=0}^{10} (-1)^n\binom{\frac{1}{2}}{n}=\frac{\binom{20}{10}}{k^{10}}$$ The result should be a rather small integer. Just use Excel and inspection.

Answer 3

It is helpful to know that multiplication of a series $A(x)=a_0+a_1x+a_2x^2+a_3x^3+\cdots$ with $\frac{1}{1-x}$ transforms the series into \begin{align*} \frac{1}{1-x}A(x)=a_0+\left(a_0+a_1\right)x+\left(a_0+a_1+a_2\right)x^2+\left(a_0+a_1+a_2+a_3\right)x^3+\cdots \end{align*} so that the coefficient of $x^{n}$ is the sum $a_0+a_1+\cdots+a_n$.

In order to find $a_0+a_1+\cdots+a_{10}$ of $\left(1-x\right)^{\frac{1}{2}}$ we look for the coefficient of $x^{10}$ in $\frac{1}{1-x}\left(1-x\right)^{\frac{1}{2}}$. We obtain \begin{align*} \frac{1}{1-x}\left(1-x\right)^{\frac{1}{2}}&=\left(1-x\right)^{-\frac{1}{2}}\tag{1}\\ &=1+\left(-\frac{1}{2}\right)(-x)+\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)\frac{(-x)^2}{2!}\\ &\qquad+\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)\frac{(-x)^{3}}{3!}\\ &\qquad+\cdots\\ &\qquad\,\,\color{blue}{+\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)\cdots\left(-\frac{1}{2}-9\right)\frac{(-x)^{10}}{10!}}\tag{2}\\ &\qquad+\cdots \end{align*}

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

We obtain from (1) and (2) \begin{align*} \color{blue}{a_0+a_1+a_2+\cdots+a_{10}}&=[x^{10}]\left(1-x\right)^{-\frac{1}{2}}\\ &=\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)\cdots\left(-\frac{19}{2}\right)\frac{(-1)^{10}}{10!}\\ &=\frac{1\cdot3\cdot5\cdots19}{2^{10}10!}\\ &=\frac{1\cdot3\cdot5\cdots19}{2^{10}10!}\cdot\frac{2\cdot4\cdot6\cdots20}{2\cdot4\cdot6\cdots20}\\ &=\frac{1\cdot2\cdot3\cdots20}{2^{10}\,10!\,2^{10}10!}\\ &\,\,\color{blue}{=\frac{1}{4^{10}}\binom{20}{10}} \end{align*} and $\color{blue}{k=4}$ follows.

The validity of the transformation can be shown via \begin{align*} &\left(a_0+a_1x+a_2x^2+a_3x^3\cdots\right)\frac{1}{1-x}\\ &\qquad=\left(a_0+a_1x+a_2x^2+a_3x^3+\cdots\right)\left(1+x+x^2+\cdots\right)\\ &\qquad=a_0+a_1x+a_2x^2+a_3x^3\cdots\\ &\qquad\qquad\ \,+a_0x+a_1x^2+a_2x^3+\cdots\\ &\qquad\qquad\qquad\quad\ +a_0x^2+a_1x^3+\cdots\\ &\qquad\qquad\qquad\qquad\qquad\;\ +\cdots\\ &\qquad=a_0+\left(a_0+a_1\right)x+\left(a_0+a_1+a_2\right)x^2+\cdots \end{align*}