On January 3, 2019, the Chang'e-4 probe successfully achieved humanity's first soft landing on the far side of the moon. A key technical challenge in achieving soft landing on the far side of the moon is maintaining communication between the ground and the probe. To solve this problem, the Chang'e-4 relay satellite ``Queqiao'' was launched, which orbits around the Earth-Moon Lagrange point $L _ { 2 }$. The $L _ { 2 }$ point is an equilibrium point located on the extension of the Earth-Moon line. Let the mass of Earth be $M _ { 1 }$, the mass of the Moon be $M _ { 2 }$, the Earth-Moon distance be $R$, and the distance from the $L _ { 2 }$ point to the Moon be $r$. According to Newton's laws of motion and the law of universal gravitation, $r$ satisfies the equation: $\frac { M _ { 1 } } { ( R + r ) ^ { 2 } } + \frac { M _ { 2 } } { r ^ { 2 } } = ( R + r ) \frac { M _ { 1 } } { R ^ { 3 } }$.
Let $\alpha = \frac { r } { R }$. Since $\alpha$ is very small, in approximate calculations $\frac { 3 \alpha ^ { 3 } + 3 \alpha ^ { 4 } + \alpha ^ { 5 } } { ( 1 + \alpha ) ^ { 2 } } \approx 3 \alpha ^ { 3 }$. Then the approximate value of $r$ is
A．$\sqrt { \frac { M _ { 2 } } { M _ { 1 } } } R$
B．$\sqrt { \frac { M _ { 2 } } { 2 M _ { 1 } } R }$
C．$\sqrt [ 3 ] { \frac { 3 M _ { 2 } } { M _ { 1 } } R }$
D．$\sqrt [ 3 ] { \frac { M _ { 2 } } { 3 M _ { 1 } } } R$

For $\alpha \in \mathbb{R}$, recall, without giving a proof, the power series expansion of $( 1 + x ) ^ { \alpha }$ on $]-1,1[$.
Justify the formula: $$\forall x \in ]-1,1[ , \quad \frac { 1 } { \sqrt { 1 - x } } = \sum _ { n = 0 } ^ { + \infty } \frac { \binom { 2 n } { n } } { 4 ^ { n } } x ^ { n }$$

Determine the power series expansion of the function $u \mapsto \sqrt{1-u}$. We will write the coefficients as a quotient of factorials and powers of 2.

Determine the power series expansion of the function $u \mapsto \sqrt{1-u}$. We will write the coefficients as a quotient of factorials and powers of 2.

Justify the existence of a sequence of real numbers $\left( a _ { n } \right) _ { n \in \mathbb { N } }$ such that $$\forall x \in ] - 1,1 \left[ , \quad \sqrt { 1 + x } = 1 + \sum _ { n = 0 } ^ { + \infty } a _ { n } x ^ { n + 1 } , \right.$$ and, for all $n \in \mathbb { N }$, express $a _ { n }$ using a binomial coefficient.

The sum of coefficients of integral powers of $x$ in the binomial expansion of $( 1 - 2 \sqrt { x } ) ^ { 50 }$ is
(1) $\frac { 1 } { 2 } \left( 2 ^ { 50 } + 1 \right)$
(2) $\frac { 1 } { 2 } \left( 3 ^ { 50 } + 1 \right)$
(3) $\frac { 1 } { 2 } \left( 3 ^ { 50 } \right)$
(4) $\frac { 1 } { 2 } \left( 3 ^ { 50 } - 1 \right)$

If the third term in the binomial expansion of $\left( 1 + x ^ { \log _ { 2 } x } \right) ^ { 5 }$ equals 2560, then a possible value of $x$ is
(1) $4 \sqrt { 2 }$
(2) $\frac { 1 } { 8 }$
(3) $2 \sqrt { 2 }$
(4) $\frac { 1 } { 4 }$

If $\left( \frac { 3 ^ { 6 } } { 4 ^ { 4 } } \right) k$ is the term, independent of $x$, in the binomial expansion of $\left( \frac { x } { 4 } - \frac { 12 } { x ^ { 2 } } \right) ^ { 12 }$, then $k$ is equal to

The sum of the coefficient of $x ^ { 2 / 3 }$ and $x ^ { - 2 / 5 }$ in the binomial expansion of $\left( x ^ { 2 / 3 } + \frac { 1 } { 2 } x ^ { - 2 / 5 } \right) ^ { 9 }$ is
(1) $21/4$
(2) $63/16$
(3) $19/4$
(4) $69/16$

If $\alpha$ satisfies the equation $x ^ { 2 } + x + 1 = 0$ and $( 1 + \alpha ) ^ { 7 } = \mathrm { A } + \mathrm { B } \alpha + \mathrm { C } \alpha ^ { 2 } , \mathrm {~A} , \mathrm {~B} , \mathrm { C } \geq 0$, then $5 ( 3 \mathrm {~A} - 2 \mathrm {~B} - \mathrm { C } )$ is equal to