O valor de $\binom{10}{3}$ é
(A) 60 (B) 90 (C) 120 (D) 150 (E) 180

The following is a proof by mathematical induction that the equality $$\sum _ { k = 0 } ^ { n } \frac { { } _ { n } \mathrm { C } _ { k } } { { } _ { n + 4 } \mathrm { C } _ { k } } = \frac { n + 5 } { 5 }$$ holds for all natural numbers $n$.
(1) When $n = 1$, $$( \text { LHS } ) = \frac { { } _ { 1 } \mathrm { C } _ { 0 } } { { } _ { 5 } \mathrm { C } _ { 0 } } + \frac { { } _ { 1 } \mathrm { C } _ { 1 } } { { } _ { 5 } \mathrm { C } _ { 1 } } = \frac { 6 } { 5 } , ( \text { RHS } ) = \frac { 1 + 5 } { 5 } = \frac { 6 } { 5 }$$ so the given equality holds.
(2) Assume that when $n = m$, the equality $$\sum _ { k = 0 } ^ { m } \frac { { } _ { m } \mathrm { C } _ { k } } { { } _ { m + 4 } \mathrm { C } _ { k } } = \frac { m + 5 } { 5 }$$ holds. When $n = m + 1$, $$\sum _ { k = 0 } ^ { m + 1 } \frac { { } _ { m + 1 } \mathrm { C } _ { k } } { { } _ { m + 5 } \mathrm { C } _ { k } } = \text { (가) } + \sum _ { k = 0 } ^ { m } \frac { { } _ { m + 1 } \mathrm { C } _ { k + 1 } } { { } _ { m + 5 } \mathrm { C } _ { k + 1 } }$$ For a natural number $l$, $${ } _ { l + 1 } \mathrm { C } _ { k + 1 } = \text { (나) } \cdot { } _ { l } \mathrm { C } _ { k } \quad ( 0 \leqq k \leqq l )$$ so $$\sum _ { k = 0 } ^ { m } \frac { { } _ { m + 1 } \mathrm { C } _ { k + 1 } } { { } _ { m + 5 } \mathrm { C } _ { k + 1 } } = \text { (다) } \cdot \sum _ { k = 0 } ^ { m } \frac { { } _ { m } \mathrm { C } _ { k } } { { } _ { m + 4 } \mathrm { C } _ { k } }$$ Therefore, $$\begin{aligned} \sum _ { k = 0 } ^ { m + 1 } \frac { { } _ { m + 1 } \mathrm { C } _ { k } } { { } _ { m + 5 } \mathrm { C } _ { k } } & = \text { (가) } + \text { (다) } \cdot \sum _ { k = 0 } ^ { m } \frac { { } _ { m } \mathrm { C } _ { k } } { { } _ { m + 4 } \mathrm { C } _ { k } } \\ & = \frac { m + 6 } { 5 } \end{aligned}$$ Thus, the given equality holds for all natural numbers $n$.
What are the correct values for (가), (나), and (다) in the above process? [3 points] $\begin{array} { l l l l } & \text { (가) } & \text { (나) } & \text { (다) } \\ \text { (1) } & 1 & \frac { l + 2 } { k + 2 } & \frac { m + 4 } { m + 4 } \end{array}$
(2) $1 \quad \frac { l + 1 } { k + 1 } \quad \frac { m + 1 } { m + 5 }$
(3) $1 \quad \frac { l + 1 } { k + 1 } \quad \frac { m + 1 } { m + 4 }$
(4) $m + 1 \quad \frac { l + 1 } { k + 1 } \quad \frac { m + 1 } { m + 5 }$
(5) $m + 1 \quad \frac { l + 2 } { k + 2 } \quad \frac { m + 1 } { m + 4 }$

The following is a proof by mathematical induction that the equality $$\sum _ { k = 0 } ^ { n } \frac { { } _ { n } \mathrm { C } _ { k } } { { } _ { n + 4 } \mathrm { C } _ { k } } = \frac { n + 5 } { 5 }$$ holds for all natural numbers $n$.

(1) When $n = 1$, $$( \text { Left side } ) = \frac { { } _ { 1 } \mathrm { C } _ { 0 } } { { } _ { 5 } \mathrm { C } _ { 0 } } + \frac { { } _ { 1 } \mathrm { C } _ { 1 } } { { } _ { 5 } \mathrm { C } _ { 1 } } = \frac { 6 } { 5 } , ( \text { Right side } ) = \frac { 1 + 5 } { 5 } = \frac { 6 } { 5 }$$ so the given equality holds.
(2) Assume that when $n = m$, the equality $$\sum _ { k = 0 } ^ { m } \frac { { } _ { m } \mathrm { C } _ { k } } { { } _ { m + 4 } \mathrm { C } _ { k } } = \frac { m + 5 } { 5 }$$ holds. When $n = m + 1$, $$\sum _ { k = 0 } ^ { m + 1 } \frac { m + 1 } { { } _ { m + 5 } \mathrm { C } _ { k } } = \text { (가) } + \sum _ { k = 0 } ^ { m } \frac { m + 1 } { { } _ { m + 5 } \mathrm { C } _ { k + 1 } }$$ For a natural number $l$, $${ } _ { l + 1 } \mathrm { C } _ { k + 1 } = \text { (나) } \cdot { } _ { l } \mathrm { C } _ { k } \quad ( 0 \leqq k \leqq l )$$ so $$\sum _ { k = 0 } ^ { m } \frac { m + 1 } { { } _ { m + 5 } \mathrm { C } _ { k + 1 } } = \text { (다) } \cdot \sum _ { k = 0 } ^ { m } \frac { { } _ { m } \mathrm { C } _ { k } } { { } _ { m + 4 } \mathrm { C } _ { k } }$$ Therefore, $$\begin{aligned} \sum _ { k = 0 } ^ { m + 1 } \frac { m + 1 } { { } _ { m + 5 } \mathrm { C } _ { k } } & = \text { (가) } + \text { (다) } \cdot \sum _ { k = 0 } ^ { m } \frac { { } _ { m } \mathrm { C } _ { k } } { { } _ { m + 4 } \mathrm { C } _ { k } } \\ & = \frac { m + 6 } { 5 } \end{aligned}$$ Thus, the given equality holds for all natural numbers $n$.
Which of the following are correct for (가), (나), and (다)? [3 points]

(가)	(나)	(다)
(1) 1	$\frac { l + 2 } { k + 2 }$	$\frac { m + 1 } { m + 4 }$
(2) 1	$\frac { l + 1 } { k + 1 }$	$\frac { m + 1 } { m + 5 }$
(3) 1	$\frac { l + 1 } { k + 1 }$	$\frac { m + 1 } { m + 4 }$
(4) $m + 1$	$\frac { l + 1 } { k + 1 }$	$\frac { m + 1 } { m + 5 }$
(5) $m + 1$	$\frac { l + 2 } { k + 2 }$	$\frac { m + 1 } { m + 4 }$

For an application $f : \mathbb{R}_+^* \rightarrow \mathbb{R}$ of class $\mathcal{C}^\infty$, we define the application $$\delta(f) : \left\{ \begin{array}{l} \mathbb{R}_+^* \rightarrow \mathbb{R} \\ x \mapsto f(x+1) - f(x) \end{array} \right.$$
For $n \in \mathbb{N}$ and $x > 0$, express $\left(\delta^n(f)\right)(x)$ using the binomial coefficients $\binom{n}{j}$ and the $f(x+j)$ (where the index $j$ belongs to $\llbracket 0, n \rrbracket$).

For every natural integer $k$ we set $$m_{k} = \frac{1}{2\pi} \int_{-2}^{2} x^{k} \sqrt{4 - x^{2}} \, \mathrm{d}x$$ Deduce that $$m_{k} = \begin{cases} C_{k/2} & \text{if } k \text{ is even} \\ 0 & \text{if } k \text{ is odd.} \end{cases}$$

For every natural integer $k$ we set $$m_{k} = \frac{1}{2\pi} \int_{-2}^{2} x^{k} \sqrt{4 - x^{2}} \, \mathrm{d}x$$ Deduce that $$m_{k} = \begin{cases} C_{k/2} & \text{if } k \text{ is even} \\ 0 & \text{if } k \text{ is odd.} \end{cases}$$

The value of $\lim_{n \to \infty} \frac{\sum_{r=0}^{n} {}^{2n}C_{2r} \times 3^{r}}{\sum_{r=0}^{n-1} {}^{2n}C_{2r+1} \times 3^{r}}$ is
(a) 0
(b) 1
(c) $\sqrt{3}$
(d) $(\sqrt{3}-1)/(\sqrt{3}+1)$

For $k \geq 1$, the value of $$\binom { n } { 0 } + \binom { n + 1 } { 1 } + \binom { n + 2 } { 2 } + \cdots + \binom { n + k } { k }$$ equals
(A) $\binom { n + k + 1 } { n + k }$
(B) $( n + k + 1 ) \binom { n + k } { n + 1 }$
(C) $\binom { n + k + 1 } { n + 1 }$
(D) $\binom { n + k + 1 } { n }$

For $k \geq 1$, the value of $\binom { n } { 0 } + \binom { n + 1 } { 1 } + \binom { n + 2 } { 2 } + \cdots + \binom { n + k } { k }$ equals
(a) $\binom { n + k + 1 } { n + k }$.
(B) $( n + k + 1 ) \binom { n + k } { n + 1 }$.
(C) $\binom { n + k + 1 } { n + 1 }$.
(D) $\binom { n + k + 1 } { n }$.

For $\mathrm { r } = 0,1 , \ldots , 10$, let $\mathrm { A } _ { \mathrm { r } } , \mathrm { B } _ { \mathrm { r } }$ and $\mathrm { C } _ { \mathrm { r } }$ denote, respectively, the coefficient of $\mathrm { x } ^ { \mathrm { r } }$ in the expansions of $( 1 + \mathrm { x } ) ^ { 10 } , ( 1 + \mathrm { x } ) ^ { 20 }$ and $( 1 + \mathrm { x } ) ^ { 30 }$. Then
$$\sum _ { r = 1 } ^ { 10 } A _ { r } \left( B _ { 10 } B _ { r } - C _ { 10 } A _ { r } \right)$$
is equal to
A) $\mathrm { B } _ { 10 } - \mathrm { C } _ { 10 }$
B) $\mathrm { A } _ { 10 } \left( \mathrm {~B} _ { 10 } ^ { 2 } - \mathrm { C } _ { 10 } \mathrm {~A} _ { 10 } \right)$
C) O
D) $\mathrm { C } _ { 10 } - \mathrm { B } _ { 10 }$

Let
$$X = \left( { } ^ { 10 } C _ { 1 } \right) ^ { 2 } + 2 \left( { } ^ { 10 } C _ { 2 } \right) ^ { 2 } + 3 \left( { } ^ { 10 } C _ { 3 } \right) ^ { 2 } + \cdots + 10 \left( { } ^ { 10 } C _ { 10 } \right) ^ { 2 }$$
where ${ } ^ { 10 } C _ { r } , r \in \{ 1,2 , \cdots , 10 \}$ denote binomial coefficients. Then, the value of $\frac { 1 } { 1430 } X$ is $\_\_\_\_$.

Suppose $$\det\left[\begin{array}{cc}\sum_{k=0}^{n} k & \sum_{k=0}^{n} {}^nC_k k^2 \\ \sum_{k=0}^{n} {}^nC_k & \sum_{k=0}^{n} {}^nC_k 3^k\end{array}\right] = 0$$ holds for some positive integer $n$. Then $\sum_{k=0}^{n} \frac{{}^nC_k}{k+1}$ equals\_\_\_\_

For nonnegative integers $s$ and $r$, let $$\binom{s}{r} = \begin{cases} \frac{s!}{r!(s-r)!} & \text{if } r \leq s \\ 0 & \text{if } r > s \end{cases}$$ For positive integers $m$ and $n$, let $$g(m, n) = \sum_{p=0}^{m+n} \frac{f(m, n, p)}{\binom{n+p}{p}}$$ where for any nonnegative integer $p$, $$f(m, n, p) = \sum_{i=0}^{p} \binom{m}{i}\binom{n+i}{p}\binom{p+n}{p-i}.$$ Then which of the following statements is/are TRUE?
(A) $g(m, n) = g(n, m)$ for all positive integers $m, n$
(B) $g(m, n+1) = g(m+1, n)$ for all positive integers $m, n$
(C) $g(2m, 2n) = 2g(m, n)$ for all positive integers $m, n$
(D) $g(2m, 2n) = (g(m, n))^{2}$ for all positive integers $m, n$

The sum of the series ${ } ^ { 20 } \mathrm { C } _ { 0 } - { } ^ { 20 } \mathrm { C } _ { 1 } + { } ^ { 20 } \mathrm { C } _ { 2 } - { } ^ { 20 } \mathrm { C } _ { 3 } + \ldots - \ldots + { } ^ { 20 } \mathrm { C } _ { 10 }$ is
(1) $- { } ^ { 20 } \mathrm { C } _ { 10 }$
(2) $\frac { 1 } { 2 } { } ^ { 20 } \mathrm { C } _ { 10 }$
(3) 0
(4) ${ } ^ { 20 } \mathrm { C } _ { 10 }$

The sum $\sum_{r=1}^{9} \frac{10!}{r!(10-r)!}$ is equal to:
(1) $2^{10} - 2$
(2) $2^{10} - 1$
(3) $2^9$
(4) $2^{10}$

The value of ${}^{21}C_1 - {}^{10}C_1 + {}^{21}C_2 - {}^{10}C_2 + {}^{21}C_3 - {}^{10}C_3 + {}^{21}C_4 - {}^{10}C_4 + \ldots + {}^{21}C_{10} - {}^{10}C_{10}$ is
(1) $2^{21} - 2^{11}$
(2) $2^{21} - 2^{10}$
(3) $2^{20} - 2^{9}$
(4) $2^{20} - 2^{10}$

The sum of the co-efficient of all odd degree terms in the expansion of $\left( x + \sqrt { x ^ { 3 } - 1 } \right) ^ { 5 } + \left( x - \sqrt { x ^ { 3 } - 1 } \right) ^ { 5 } , ( x > 1 )$ is
(1) 2
(2) - 1
(3) 0
(4) 1

If $\sum _ { i = 1 } ^ { 20 } \left( \frac { { } ^ { 20 } C _ { i - 1 } } { { } ^ { 20 } C _ { i } + { } ^ { 20 } C _ { i - 1 } } \right) ^ { 3 } = \frac { k } { 21 }$, then $k$ equals
(1) 200
(2) 100
(3) 50
(4) 400

The sum of the co-efficients of all even degree terms in $x$ in the expansion of $\left(x + \sqrt{x^3 - 1}\right)^6 + \left(x - \sqrt{x^3 - 1}\right)^6$, $x > 1$ is equal to
(1) 26
(2) 32
(3) 24
(4) 29

The value of $-{ } ^ { 15 } C _ { 1 } + 2 \cdot { } ^ { 15 } C _ { 2 } - 3 \cdot { } ^ { 15 } C _ { 3 } + \ldots - 15 \cdot { } ^ { 15 } C _ { 15 } + { } ^ { 14 } C _ { 1 } + { } ^ { 14 } C _ { 3 } + { } ^ { 14 } C _ { 5 } + \ldots + { } ^ { 14 } C _ { 11 }$ is equal to
(1) $2 ^ { 14 }$
(2) $2 ^ { 13 } - 13$
(3) $2 ^ { 16 } - 1$
(4) $2 ^ { 13 } - 14$

If ${ } ^ { 20 } \mathrm { C } _ { \mathrm { r } }$ is the co-efficient of $x ^ { \mathrm { r } }$ in the expansion of $( 1 + x ) ^ { 20 }$, then the value of $\sum _ { \mathrm { r } = 0 } ^ { 20 } \mathrm { r } ^ { 2 } \left( { } ^ { 20 } \mathrm { C } _ { \mathrm { r } } \right)$ is equal to:
(1) $420 \times 2 ^ { 18 }$
(2) $380 \times 2 ^ { 18 }$
(3) $380 \times 2^{19}$
(4) $420 \times 2 ^ { 19 }$

Let $[ x ]$ denote greatest integer less than or equal to $x$. If for $n \in N , \left( 1 - x + x ^ { 3 } \right) ^ { n } = \sum _ { j = 0 } ^ { 3 n } a _ { j } x ^ { j }$, then $\sum _ { j = 0 } ^ { \left[ \frac { 3 n } { 2 } \right] } a _ { 2 j } + 4 \sum _ { j = 0 } ^ { \left[ \frac { 3 n - 1 } { 2 } \right] } a _ { 2 j + 1 }$ is equal to :
(1) 2
(2) $2 ^ { n - 1 }$
(3) 1
(4) $n$

The value of $\sum _ { r = 0 } ^ { 6 } \left( { } ^ { 6 } C _ { r } \cdot { } ^ { 6 } C _ { 6 - r } \right)$ is equal to:
(1) 1124
(2) 1324
(3) 1024
(4) 924

$\sum _ { i , j = 0 , i \neq j } ^ { n } { } ^ { n } C _ { i } { } ^ { n } C _ { j }$ is equal to
(1) $2 ^ { 2 n } - { } ^ { 2 n } C _ { n }$
(2) $2 ^ { 2 n - 1 } - { } ^ { 2 n - 1 } C _ { n - 1 }$
(3) $2 ^ { 2 n } - \frac { 1 } { 2 } { } ^ { 2 n } C _ { n }$
(4) $2 ^ { n - 1 } + { } ^ { 2 n - 1 } C _ { n }$

If $1 + \left( 2 + { } ^ { 49 } C _ { 1 } + { } ^ { 49 } C _ { 2 } + \ldots + { } ^ { 49 } C _ { 49 } \right) \left( { } ^ { 50 } C _ { 2 } + { } ^ { 50 } C _ { 4 } + \ldots + { } ^ { 50 } C _ { 50 } \right)$ is equal to $2 ^ { n } \cdot m$, where $m$ is odd, then $n + m$ is equal to $\_\_\_\_$.