Newton binomial expansion

Find the coefficient for ${x^{11}}$ in ${\sqrt{1+x}}$

The generic formula is

$$\sqrt{1+x}=(1+x)^\frac{1}{2}=1+\frac{1}{2}x+\frac{\frac{1}{2}(\frac{1}{2}-1)}{2}x^2\dots$$

Solution of expansion of coeficient for ${x^{11}}$:

$${\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)(\frac{1}{2}-3)(\frac{1}{2}-4)(\frac{1}{2}-5)(\frac{1}{2}-6)(\frac{1}{2}-7)(\frac{1}{2}-8)(\frac{1}{2}-9)(\frac{1}{2}-10)}{11!}}$$

$$={\frac{\frac{654729075}{2^{11}}}{11!}}x^{11}$$

My question is: is this correct?

• yes, it is correct – G Cab Jan 13 '18 at 10:03

$(1+x)^a=1+ax+...+\dfrac{a(a-1)\cdots(a-(n-1))}{n!}x^n+o(x^n)$
Let show by induction $\sqrt{1+x}=\sum\limits_{k=0}^n a_kx^k+o(x^n)$ with $$a_n=\dfrac{(-1)^{n-1}}{(2n-1)}\dfrac{\binom{2n}{n}}{4^n}$$
$a_0=\dfrac{-1}{-1}=0\quad\checkmark$
$a_1=\dfrac{2}{4}=\frac 12\quad\checkmark$
$a_{n+1}=a_n\times\dfrac{(\frac 12-n)}{(n+1)}=\dfrac{(-1)^{n-1}(2n)!(-\frac 12)(2n-1)}{(2n-1)n!n!4^n(n+1)}=\dfrac{(-1)^n(2n+2)!}{(n+1)!n!(2n+1)(2n+2)4^n\times 2}=\dfrac{(-1)^n(2n+2)!}{4^{n+1}(n+1)!(n+1)!(2n+1)}\quad\checkmark$
So $a_{11}=\dfrac{\binom{22}{11}}{4^{11}\times 21}=\dfrac{4199}{524288}$, this is also the result you got!