The sum $$S = 1 + 111 + 11111 + \cdots + \underbrace{11\cdots1}_{2k+1}$$ is equal to . . . . . . .

An arithmetic sequence $\left\{ a _ { n } \right\}$ with all positive terms and equal first term and common difference satisfies $$\sum _ { k = 1 } ^ { 15 } \frac { 1 } { \sqrt { a _ { k } } + \sqrt { a _ { k + 1 } } } = 2$$ What is the value of $a _ { 4 }$? [3 points]
(1) 6
(2) 7
(3) 8
(4) 9
(5) 10

For an arithmetic sequence $\{a_n\}$ with nonzero common difference, $$|a_6| = a_8, \quad \sum_{k=1}^{5} \frac{1}{a_k a_{k+1}} = \frac{5}{96}$$ Find the value of $\sum_{k=1}^{15} a_k$. [4 points]
(1) 60
(2) 65
(3) 70
(4) 75
(5) 80

11. The sequence $\left\{ a _ { n } \right\}$ satisfies $a _ { 1 } = 1$ and $a _ { n + 1 } - a _ { n } = n + 1 \quad \left( n \in N ^ { * } \right)$, then the sum of the first 10 terms of the sequence $\left\{ \frac { 1 } { a _ { n } } \right\}$ is $\_\_\_\_$.

15. For an arithmetic sequence $\left\{ a _ { n } \right\}$ with sum of first $n$ terms $S _ { n }$, if $a _ { 3 } = \frac { 3 } { 2 } , S _ { 4 } = 10$, then $\sum_{k=1}^{n} \frac { 1 } { S _ { k } } = $ ______ [Figure]
III. Solving Problems: Questions 16-23 [Figure] [Figure]

Recall Stirling's formula. Deduce an asymptotic equivalent of $C _ { n }$ as $n$ tends to $+ \infty$.

From the previous question, recover the result of questions 11 and 16.

Let $7 \overbrace { 5 \cdots 5 } ^ { r } 7$ denote the $( r + 2 )$ digit number where the first and the last digits are 7 and the remaining $r$ digits are 5. Consider the sum $S = 77 + 757 + 7557 + \cdots + 7 \overbrace { 5 \cdots 5 } ^ { 98 } 7$. If $S = \frac { 7 \overbrace { 5 \cdots 5 } ^ { 99 } 7 + m } { n }$, where $m$ and $n$ are natural numbers less than 3000, then the value of $m + n$ is

The sum of first 20 terms of the sequence $0.7, 0.77, 0.777, \ldots\ldots$, is:
(1) $\frac{7}{81}\left(179 + 10^{-20}\right)$
(2) $\frac{7}{9}\left(99 + 10^{-20}\right)$
(3) $\frac{7}{81}\left(179 - 10^{-20}\right)$
(4) $\frac{7}{9}\left(99 - 10^{-20}\right)$

If the sum $\frac { 3 } { 1 ^ { 2 } } + \frac { 5 } { 1 ^ { 2 } + 2 ^ { 2 } } + \frac { 7 } { 1 ^ { 2 } + 2 ^ { 2 } + 3 ^ { 2 } } + \ldots + $ up to 20 terms is equal to $\frac { k } { 21 }$, then $k$ is equal to
(1) 240
(2) 120
(3) 60
(4) 180

The sum of first 9 terms of the series $\frac { 1 ^ { 3 } } { 1 } + \frac { 1 ^ { 3 } + 2 ^ { 3 } } { 1 + 3 } + \frac { 1 ^ { 3 } + 2 ^ { 3 } + 3 ^ { 3 } } { 1 + 3 + 5 } + \ldots$ is
(1) 192
(2) 71
(3) 96
(4) 142

Let $\frac { 1 } { x _ { 1 } } , \frac { 1 } { x _ { 2 } } , \ldots , \frac { 1 } { x _ { n } } \left( x _ { i } \neq 0 \right.$ for $\left. i = 1,2 , \ldots , n \right)$ be in A.P. such that $x _ { 1 } = 4$ and $x _ { 21 } = 20$. If $n$ is the least positive integer for which $x _ { n } > 50$, then $\sum _ { i = 1 } ^ { n } \left( \frac { 1 } { x _ { i } } \right)$ is equal to
(1) 3
(2) $\frac { 1 } { 8 }$
(3) $\frac { 13 } { 4 }$
(4) $\frac { 13 } { 8 }$

The sum of the following series $1 + 6 + \frac{9\left(1^2 + 2^2 + 3^2\right)}{7} + \frac{12\left(1^2 + 2^2 + 3^2 + 4^2\right)}{9} + \frac{15\left(1^2 + 2^2 + \ldots + 5^2\right)}{11} + \ldots$ up to 15 terms, is:
(1) 7520
(2) 7510
(3) 7830
(4) 7820

Let $a _ { 1 } , a _ { 2 } , \ldots , a _ { 21 }$ be an A.P. such that $\sum _ { n = 1 } ^ { 20 } \frac { 1 } { a _ { n } a _ { n + 1 } } = \frac { 4 } { 9 }$. If the sum of this A.P. is 189 , then $\mathrm { a } _ { 6 } \mathrm { a } _ { 16 }$ is equal to :
(1) 57
(2) 48
(3) 36
(4) 72

The sum of 10 terms of the series $\frac { 3 } { 1 ^ { 2 } \times 2 ^ { 2 } } + \frac { 5 } { 2 ^ { 2 } \times 3 ^ { 2 } } + \frac { 7 } { 3 ^ { 2 } \times 4 ^ { 2 } } + \ldots$ is :
(1) $\frac { 143 } { 144 }$
(2) $\frac { 99 } { 100 }$
(3) 1
(4) $\frac { 120 } { 121 }$

If $\frac { 1 } { ( 20 - a ) ( 40 - a ) } + \frac { 1 } { ( 40 - a ) ( 60 - a ) } + \ldots\ldots + \frac { 1 } { ( 180 - a ) ( 200 - a ) } = \frac { 1 } { 256 }$, then the maximum value of $a$ is
(1) 198
(2) 202
(3) 212
(4) 218

$\frac { 2 ^ { 3 } - 1 ^ { 3 } } { 1 \times 7 } + \frac { 4 ^ { 3 } - 3 ^ { 3 } + 2 ^ { 3 } - 1 ^ { 3 } } { 2 \times 11 } + \frac { 6 ^ { 3 } - 5 ^ { 3 } + 4 ^ { 3 } - 3 ^ { 3 } + 2 ^ { 3 } - 1 ^ { 3 } } { 3 \times 15 } + \ldots \ldots + \frac { 30 ^ { 3 } - 29 ^ { 3 } + 28 ^ { 3 } - 27 ^ { 3 } + \ldots + 2 ^ { 3 } - 1 ^ { 3 } } { 15 \times 63 }$ is equal to $\_\_\_\_$ .

The sum to 10 terms of the series $\frac{1}{1 + 1^2 + 1^4} + \frac{2}{1 + 2^2 + 2^4} + \frac{3}{1 + 3^2 + 3^4} + \ldots$ is:
(1) $\frac{59}{111}$
(2) $\frac{55}{111}$
(3) $\frac{56}{111}$
(4) $\frac{58}{111}$

Suppose f is a function satisfying $\mathrm { f } ( \mathrm { x } + \mathrm { y } ) = \mathrm { f } ( \mathrm { x } ) + \mathrm { f } ( \mathrm { y } )$ for all $\mathrm { x } , \mathrm { y } \in \mathbb { N }$ and $\mathrm { f } ( 1 ) = \frac { 1 } { 5 }$. If $\sum _ { n = 1 } ^ { m } \frac { f ( n ) } { n ( n + 1 ) ( n + 2 ) } = \frac { 1 } { 12 }$ then m is equal to $\_\_\_\_$ .

Let $a_{1} = 1, a_{2}, a_{3}, a_{4}, \ldots$ be consecutive natural numbers. Then $\tan^{-1}\left(\frac{1}{1 + a_{1}a_{2}}\right) + \tan^{-1}\left(\frac{1}{1 + a_{2}a_{3}}\right) + \ldots + \tan^{-1}\left(\frac{1}{1 + a_{2021}a_{2022}}\right)$ is equal to
(1) $\frac{\pi}{4} - \cot^{-1}(2022)$
(2) $\cot^{-1}(2022) - \frac{\pi}{4}$
(3) $\tan^{-1}(2022) - \frac{\pi}{4}$
(4) $\frac{\pi}{4} - \tan^{-1}(2022)$

Let $a_1, a_2, \ldots, a_n$ be in A.P. If $a_5 = 2a_7$ and $a_{11} = 18$, then $12\left(\frac{1}{\sqrt{a_{10}} + \sqrt{a_{11}}} + \frac{1}{\sqrt{a_{11}} + \sqrt{a_{12}}} + \ldots + \frac{1}{\sqrt{a_{17}} + \sqrt{a_{18}}}\right)$ is equal to $\underline{\hspace{1cm}}$.

The value of $\frac { 1 \times 2 ^ { 2 } + 2 \times 3 ^ { 2 } + \ldots + 100 \times ( 101 ) ^ { 2 } } { 1 ^ { 2 } \times 2 + 2 ^ { 2 } \times 3 + \ldots + 100 ^ { 2 } \times 101 }$ is
(1) $\frac { 32 } { 31 }$
(2) $\frac { 31 } { 30 }$
(3) $\frac { 306 } { 305 }$
(4) $\frac { 305 } { 301 }$

If $\left( \frac { 1 } { \alpha + 1 } + \frac { 1 } { \alpha + 2 } + \ldots \ldots + \frac { 1 } { \alpha + 1012 } \right) - \left( \frac { 1 } { 2 \cdot 1 } + \frac { 1 } { 4 \cdot 3 } + \frac { 1 } { 6 \cdot 5 } + \ldots . + \frac { 1 } { 2024 \cdot 2023 } \right) = \frac { 1 } { 2024 }$, then $\alpha$ is equal to $\_\_\_\_$

$$\sum _ { n = 5 } ^ { 14 } \frac { 1 } { 1 + 2 + \cdots + n }$$
What is the value of this sum?
A) $\frac { 1 } { 3 }$
B) $\frac { 2 } { 3 }$
C) $\frac { 3 } { 5 }$
D) $\frac { 2 } { 15 }$
E) $\frac { 4 } { 15 }$

$$\left( \sum _ { k = 1 } ^ { 9 } k \right) \cdot \left( \sum _ { n = 1 } ^ { 8 } \frac { 1 } { n ( n + 1 ) } \right)$$
What is the result of this operation?
A) 27
B) 30
C) 32
D) 36
E) 40