1. ## coefficient

in the expansion of $\left ( x-\sqrt{x}-1 \right )^{7}$
wath is the coefficient of $x^{\frac{5}{2}}$

2. Hello, dapore!

$\displaystyle \text{In the expansion of: }\:\left( x-\sqrt{x}-1 \right)^{7},\;\text{ what is the coefficient of }\,x^{\frac{5}{2}}\:?$

$\displaystyle \text{The only terms containing }\,x^{\frac{5}{2}}\,\text{ are: }\;\left(x\right)^2\!\left(\text{-}x^{\frac{1}{2}}\right)\!\left(\text{-}1\right)^4,\;\;\left(x\right)^1\!\left(\text{-}x^{\frac{1}{2}}\right)^3\!\left(\text{-}1\right)^3,$ . $\displaystyle \left(x\right)^0\!\left(x^{\frac{1}{2}}\right)^5\! \left(\text{-}1\right)^2$

We have: .$\displaystyle {7\choose2,1,4}x^2\left(\text{-}x^{\frac{1}{2}}\right)(\text{-}1)^4 \;=\;105\left(-x^{\frac{5}{2}}\right)$

. . . . . . . .$\displaystyle {7\choose1,3,3}x\left(\text{-}x^{\frac{1}{2}}\right)^3(\text{-}1)^3 \;=\;140\left(+x^{\frac{5}{2}}\right)$

. . . . . . . .$\displaystyle {7\choose0,5,2}x^0\left(\text{-}x^{\frac{1}{2}}\right)^5(\text{-}1)^2 \;=\; 21\left(-x^{\frac{5}{2}}\right)$

Therefore: .$\displaystyle -105x^{\frac{5}{2}} + 140x^{\frac{5}{2}} - 21x^{\frac{5}{2}} \;\;=\;\; {\color{blue}14}\,x^{\frac{5}{2}}$

3. Originally Posted by Soroban
Hello, dapore!

$\displaystyle \text{The only terms containing }\,x^{\frac{5}{2}}\,\text{ are: }\;\left(x\right)^2\!\left(\text{-}x^{\frac{1}{2}}\right)\!\left(\text{-}1\right)^4,\;\;\left(x\right)^1\!\left(\text{-}x^{\frac{1}{2}}\right)^3\!\left(\text{-}1\right)^3,$ . $\displaystyle \left(x\right)^0\!\left(x^{\frac{1}{2}}\right)^5\! \left(\text{-}1\right)^2$

We have: .$\displaystyle {7\choose2,1,4}x^2\left(\text{-}x^{\frac{1}{2}}\right)(\text{-}1)^4 \;=\;105\left(-x^{\frac{5}{2}}\right)$

. . . . . . . .$\displaystyle {7\choose1,3,3}x\left(\text{-}x^{\frac{1}{2}}\right)^3(\text{-}1)^3 \;=\;140\left(+x^{\frac{5}{2}}\right)$

. . . . . . . .$\displaystyle {7\choose0,5,2}x^0\left(\text{-}x^{\frac{1}{2}}\right)^5(\text{-}1)^2 \;=\; 21\left(-x^{\frac{5}{2}}\right)$

Therefore: .$\displaystyle -105x^{\frac{5}{2}} + 140x^{\frac{5}{2}} - 21x^{\frac{5}{2}} \;\;=\;\; {\color{blue}14}\,x^{\frac{5}{2}}$
Thank you, thank you, thank you
But how we calculated

$\binom{7}{2,1,4}=105?$

4. Originally Posted by dapore
Thank you, thank you, thank you
But how we calculated

$\binom{7}{2,1,4}=105?$
That expression is equal to $\displaystyle \frac{7!}{2!\,1!\,4!}$

I believe it comes from the coefficients of the binomial expansion