1
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ yes, it is correct $\endgroup$ – G Cab Jan 13 '18 at 10:03
1
$\begingroup$

$(1+x)^a=1+ax+...+\dfrac{a(a-1)\cdots(a-(n-1))}{n!}x^n+o(x^n)$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.