Trắc nghiệm Tổng hợp Toán năm 2024 có đáp án - Phần 2

\[0 \in \mathbb{Z}\] vì \[\mathbb{Z} = \left\{ {...; - 1;0;1;...} \right\}.\]

Ư(0) = {ℕ^*}.

0,03kg = 0,03.1000 = 3 g.

0,5 m = 50 cm.

Lời giải: \[0,75 = \frac{{75}}{{100}} = \frac{3}{4}\]

\[0,009 = \frac{9}{{1000}}.\]

Số 0,089 đọc là “Không phẩy không trăm tám mươi chín”.

\[1,05 = 1\frac{1}{{20}} = \frac{{21}}{{20}}.\]

\[1 + \frac{1}{{1 + 2}} + \frac{1}{{1 + 2 + 3}} + \frac{1}{{1 + 2 + 3 + 4}} + ... + \frac{1}{{1 + 2 + 3 + 4 + ... + 50}}\]

\[ = 1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{{10}} + ... + \frac{1}{{1\,\,275}}\]

\[ = \frac{2}{2} + \frac{2}{6} + \frac{2}{{12}} + \frac{2}{{20}} + ... + \frac{2}{{2\,\,550}}\]

\[ = \frac{2}{{1 \cdot 2}} + \frac{2}{{2 \cdot 3}} + \frac{2}{{3 \cdot 4}} + \frac{2}{{4 \cdot 5}} + ... + \frac{2}{{50 \cdot 51}}\]

\[ = 2 \cdot \left( {\frac{1}{{1 \cdot 2}} + \frac{1}{{2 \cdot 3}} + \frac{1}{{3 \cdot 4}} + \frac{1}{{4 \cdot 5}} + ... + \frac{1}{{50 \cdot 51}}} \right)\]

\[ = 2 \cdot \left( {1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + ... + \frac{1}{{50}} - \frac{1}{{51}}} \right)\]

\[ = 2 \cdot \left( {1 - \frac{1}{{51}}} \right)\]

\[ = 2 \cdot \frac{{50}}{{51}} = \frac{{100}}{{51}}.\]

\[\frac{1}{{1 + 2}} + \frac{1}{{1 + 2 + 3}} + \frac{1}{{1 + 2 + 3 + 4}} + ... + \frac{1}{{1 + 2 + 3 + 4 + ... + 50}}\]

\[ = \frac{1}{3} + \frac{1}{6} + \frac{1}{{10}} + ... + \frac{1}{{1\,\,275}}\]

\[ = \frac{2}{6} + \frac{2}{{12}} + \frac{2}{{20}} + ... + \frac{2}{{2\,\,550}}\]

\[ = \frac{2}{{2 \cdot 3}} + \frac{2}{{3 \cdot 4}} + \frac{2}{{4 \cdot 5}} + ... + \frac{2}{{50 \cdot 51}}\]

\[ = 2 \cdot \left( {\frac{1}{{2 \cdot 3}} + \frac{1}{{3 \cdot 4}} + \frac{1}{{4 \cdot 5}} + ... + \frac{1}{{50 \cdot 51}}} \right)\]

\[ = 2 \cdot \left( {\frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + ... + \frac{1}{{50}} - \frac{1}{{51}}} \right)\]

\[ = 2 \cdot \left( {\frac{1}{2} - \frac{1}{{51}}} \right)\]

\[ = 2 \cdot \frac{{49}}{{102}} = \frac{{49}}{{51}}.\]

\[\left( {1 + \frac{1}{2}} \right) \times \left( {1 + \frac{1}{3}} \right) \times \left( {1 + \frac{1}{4}} \right) \times ... \times \left( {1 + \frac{1}{{2013}}} \right)\]

\[ = \frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times ... \times \frac{{2014}}{{2013}}\]

\[ = \frac{{2014}}{2} = 1007.\]

\[\left( {1 - \frac{1}{2}} \right) \times \left( {1 - \frac{1}{3}} \right) \times \left( {1 - \frac{1}{4}} \right) \times \left( {1 - \frac{1}{5}} \right) \times \left( {1 - \frac{1}{6}} \right) \times \left( {1 - \frac{1}{7}} \right)\]

\[ = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times \frac{5}{6} \times \frac{6}{7} = \frac{1}{7}.\]

\[A = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + ... + \frac{1}{{49}} - \frac{1}{{50}}\]

\[ = \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{{49}} - \frac{1}{{50}}\]

\[ = \frac{5}{6} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + ... + \frac{1}{{49}} - \frac{1}{{50}}\]

\[ = \frac{5}{6} + \left( {\frac{1}{5} - \frac{1}{4}} \right) + \left( {\frac{1}{7} - \frac{1}{6}} \right) + ... + \left( {\frac{1}{{49}} - \frac{1}{{48}}} \right) + \left( { - \frac{1}{{50}}} \right)\]

Mà \[\frac{1}{5} - \frac{1}{4} < 0\]

\[\frac{1}{7} - \frac{1}{6} < 0\]

...

\[\frac{1}{{49}} - \frac{1}{{48}} < 0\]

\[ - \frac{1}{{50}} < 0\]

Do đó \[A < \frac{5}{6}.\]

\[A = \left( {1 - \frac{1}{{{2^2}}}} \right) \times \left( {1 - \frac{1}{{{3^2}}}} \right) \times \left( {1 - \frac{1}{{{4^2}}}} \right) \times ... \times \left( {1 - \frac{1}{{{{30}^2}}}} \right)\]

\[ = \left[ {{1^2} - {{\left( {\frac{1}{2}} \right)}^2}} \right] \cdot \left[ {{1^2} - {{\left( {\frac{1}{3}} \right)}^2}} \right] \cdot ... \cdot \left[ {{1^2} - {{\left( {\frac{1}{{30}}} \right)}^2}} \right]\]

\[ = \left( {1 - \frac{1}{2}} \right)\left( {1 + \frac{1}{2}} \right)\left( {1 - \frac{1}{3}} \right)\left( {1 + \frac{1}{3}} \right) \cdot \cdot \cdot \left( {1 - \frac{1}{{30}}} \right)\left( {1 + \frac{1}{{30}}} \right)\]

\[ = \frac{1}{2} \cdot \frac{3}{2} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \cdot \cdot \frac{{29}}{{30}} \cdot \frac{{31}}{{30}}\]

\[ = \frac{1}{2} \cdot \frac{{31}}{{30}} = \frac{{31}}{{60}}.\]

1,2 + 2,2 + 3,3 + 4,4 + 5,5 + 6,6 + 7,7 + 8,8 + 9,9

= (1,1 + 9,9) + (2,2 + 8,8) + (3,3 + 7,7) + (4,4 + 6,6) + 5,5

= 11 + 11 + 11 + 11 + 5,5

= 49,5.

Ta có:

\[\frac{1}{{1 + 2 + 3}} + \frac{1}{{1 + 2 + 3 + 4}} + ... + \frac{1}{{1 + 2 + 3 + 4 + ... + 59}}\]

\[ = \frac{1}{{\frac{{3 \cdot 4}}{2}}} + \frac{1}{{\frac{{4 \cdot 5}}{2}}} + ... + \frac{1}{{\frac{{59 \cdot 60}}{2}}}\]

\[ = \frac{2}{{3 \cdot 4}} + \frac{2}{{4 \cdot 5}} + ... + \frac{2}{{59 \cdot 60}}\]

\[ = 2 \cdot \left( {\frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + ... + \frac{1}{{59}} - \frac{1}{{60}}} \right)\]

\[ = 2 \cdot \left( {\frac{1}{3} - \frac{1}{{60}}} \right)\]

\[ = 2 \cdot \frac{{19}}{{60}}\]

\[ = \frac{{38}}{{60}}\]

Mà \[\frac{{38}}{{60}} < \frac{{40}}{{60}} = \frac{2}{3}\]

Vậy \[M < \frac{2}{3}.\]

\[A = \left( {1 - \frac{1}{2}} \right) \times \left( {1 - \frac{1}{3}} \right) \times \left( {1 - \frac{1}{4}} \right) \times ...\left( {1 - \frac{1}{{2014}}} \right) \times \left( {1 - \frac{1}{{2015}}} \right)\]

\[ = \frac{1}{2} \cdot \frac{2}{3} \cdot ... \cdot \frac{{2014}}{{2015}} = \frac{1}{{2015}}.\]

\[A = \left( {1 - \frac{1}{2}} \right) \times \left( {1 - \frac{1}{3}} \right) \times \left( {1 - \frac{1}{4}} \right) \times ...\left( {1 - \frac{1}{{100}}} \right)\]

\[ = \frac{1}{2} \cdot \frac{2}{3} \cdot ... \cdot \frac{{99}}{{100}} = \frac{1}{{100}}.\]

\[A = 1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{{10}} + \frac{1}{{15}} + ... + \frac{1}{{120}}\]

\[ = \frac{2}{2} \cdot \left( {\frac{1}{3} + \frac{1}{6} + \frac{1}{{10}} + \frac{1}{{15}} + ... + \frac{1}{{120}}} \right)\]

\[ = 2\left( {\frac{1}{6} + \frac{1}{{12}} + \frac{1}{{20}} + \frac{1}{{30}} + ... + \frac{1}{{240}}} \right)\]

\[ = 2\left( {\frac{1}{{2 \cdot 3}} + \frac{1}{{3 \cdot 4}} + \frac{1}{{4 \cdot 5}} + \frac{1}{{5 \cdot 6}} + ... + \frac{1}{{15 \cdot 16}}} \right)\]

\[ = 2 \cdot \left( {\frac{1}{2} - \frac{1}{{16}}} \right)\]

\[ = 2 \cdot \frac{7}{{16}} = \frac{7}{8}.\]

\[\frac{1}{{1 + a}} + \frac{1}{{1 + b}} + \frac{1}{{1 + c}} = 2\]

\[ \Rightarrow \frac{1}{{a + 1}} = 1 - \frac{1}{{b + 1}} + 1 - \frac{1}{{c + 1}} = \frac{b}{{b + 1}} + \frac{c}{{c + 1}} \ge 2\sqrt {\frac{{bc}}{{\left( {b + 1} \right)\left( {c + 1} \right)}}} \]

Tương tự ta có:

\[ \Rightarrow \frac{1}{{b + 1}} \ge 2\sqrt {\frac{{ac}}{{\left( {a + 1} \right)\left( {c + 1} \right)}}} ;\frac{1}{{c + 1}} \ge 2\sqrt {\frac{{ab}}{{\left( {a + 1} \right)\left( {b + 1} \right)}}} \]

Nhân vế với vế các bất đẳng thức trên

\[ \Rightarrow \frac{1}{{\left( {a + 1} \right)\left( {b + 1} \right)\left( {c + 1} \right)}} \ge 8\sqrt {\frac{{{a^2}{b^2}{c^2}}}{{{{\left( {a + 1} \right)}^2}{{\left( {b + 1} \right)}^2}{{\left( {c + 1} \right)}^2}}}} = 8 \cdot \frac{{abc}}{{\left( {a + 1} \right)\left( {b + 1} \right)\left( {c + 1} \right)}}\]

\[ \Rightarrow 1 \le 8abc\]

\[ \Rightarrow abc \le \frac{1}{8}\]

Đẳng thức xảy ra khi \[a = b = c = \frac{1}{2}\]

Vậy giá trị lớn nhất của Q là \[\frac{1}{8}\] khi \[a = b = c = \frac{1}{2}\]

\[A = \frac{1}{{1 \cdot 3}} + \frac{1}{{3 \cdot 5}} + \frac{1}{{5 \cdot 7}} + ... + \frac{1}{{95 \cdot 97}} + \frac{1}{{97 \cdot 99}}\]

\[ = \frac{1}{2} \cdot \left( {\frac{2}{{1 \cdot 3}} + \frac{2}{{3 \cdot 5}} + ... + \frac{2}{{97 \cdot 99}}} \right)\]

\[ = \frac{1}{2} \cdot \left( {1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + ... + \frac{1}{{97}} - \frac{1}{{99}}} \right)\]

\[ = \frac{1}{2} \cdot \left( {1 - \frac{1}{{99}}} \right)\]

\[ = \frac{1}{2} \cdot \frac{{98}}{{99}} = \frac{{49}}{{99}}.\]

\[\frac{1}{{10}}ha = 1000{m^2}\]

1 phút = 60 giây

Nên \[\frac{1}{{10}}\] phút \[ = 60 \times \frac{1}{{10}} = 6\] giây.

Ta có: 1ha = 10 000m²

\[\frac{1}{{100}}ha = \frac{1}{{100}} \cdot 10000 = 100{m^2}\]

Ta có: 1ha = 10 000m²

\[\frac{1}{{10}}ha = \frac{1}{{10}} \cdot 10000 = 1000{m^2}\]

1 giờ = 60 phút

Ta có:

60 : 12 = 5 phút

Đáp số : \[\frac{1}{{12}}\] giờ = 5 phút

\[1,15 = \frac{{115}}{{100}} = \frac{{23}}{{20}}.\]

\[\frac{1}{{15}} + \frac{4}{{15}} + \frac{5}{{18}} + \frac{7}{{18}}\]

\[ = \left( {\frac{1}{{15}} + \frac{4}{{15}}} \right) + \left( {\frac{5}{{18}} + \frac{7}{{18}}} \right)\]

\[ = \frac{5}{{15}} + \frac{{12}}{{18}}\]

\[ = \frac{1}{3} + \frac{2}{3}\]

\[ = \frac{3}{3} = 1\]

Đáp án đúng là: A

Ta có: tỉ lệ 1 : 150 000 thuộc loại lớn.

Ta có: 1 = 1

16 = 2⁴

Vậy ƯCLN(1, 16) = 1.

\[\frac{1}{{1 \cdot 2}} + \frac{1}{{2 \cdot 3}} + \frac{1}{{3 \cdot 4}} + \frac{1}{{4 \cdot 5}} + \frac{1}{{5 \cdot 6}} + \frac{1}{{6 \cdot 7}}\]

\[ = \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{6} - \frac{1}{7}\]

\[ = 1 - \frac{1}{7} = \frac{6}{7}.\]

Đặt \[A = {\left( {\frac{1}{2}} \right)^0} + {\left( {\frac{1}{2}} \right)^1} + ... + {\left( {\frac{1}{2}} \right)^{10}}\]

\[ = 1 + \frac{1}{2} + ... + \frac{1}{{{2^{10}}}}\]

\[2A = 2\left( {1 + \frac{1}{2} + ... + \frac{1}{{{2^{10}}}}} \right)\]

\[2A = 2 + 1 + ... + \frac{1}{{{2^9}}}\]

\[2A - A = \left( {2 + 1 + ... + \frac{1}{{{2^9}}}} \right) - \left( {1 + \frac{1}{2} + ... + \frac{1}{{{2^{10}}}}} \right)\]

\[A = 2 - \frac{1}{{{2^{10}}}}\]

Ta có S = 1 + 2 + ...+ n

Dãy số trên là dãy số cách đều với khoảng cách là: 2 ‒ 1 = 1

Số số hạng của dãy số trên là: (n ‒ 1) : 1 + 1 = n

Áp dụng công thức tính tổng của dãy số cách đều ta có tổng trên là:

S = (n + 1). n : 2

Áp dụng công thức tính tổng S trên vào biểu thức A ta có:

\[A = \frac{1}{{\left( {2 + 1} \right) \times 2:2}} + \frac{1}{{\left( {3 + 1} \right) \times 3:2}} + ... + \frac{1}{{\left( {2020 + 1} \right) \times 2020:2}}\]

\[ = \frac{1}{{2 \times 3:2}} + \frac{1}{{3 \times 4:2}} + \frac{1}{{4 \times 5:2}} + ... + \frac{1}{{2020 \times 2021:2}}\]

\[ = \frac{2}{{2 \times 3}} + \frac{2}{{3 \times 4}} + \frac{2}{{4 \times 5}} + ... + \frac{2}{{2020 \times 2021}}\]

\[ = 2 \times \left( {\frac{1}{{2 \times 3}} + \frac{1}{{3 \times 4}} + \frac{1}{{4 \times 5}} + ... + \frac{1}{{2020 \times 2021}}} \right)\]

\[ = 2 \times \left( {\frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{4} - \frac{1}{5} + ... + \frac{1}{{2020}} - \frac{1}{{2021}}} \right)\]

\[ = 2 \cdot \left( {\frac{1}{2} - \frac{1}{{2021}}} \right)\]

\[ = 1 - \frac{2}{{2021}}\]

\[ = \frac{{2021 - 2}}{{2021}}\]

\[ = \frac{{2019}}{{2021}}\]

Ta có S = 1 + 2 + ...+ n

Dãy số trên là dãy số cách đều với khoảng cách là: 2 ‒ 1 = 1

Số số hạng của dãy số trên là: (n ‒ 1) : 1 + 1 = n

Áp dụng công thức tính tổng của dãy số cách đều ta có tổng trên là:

S = (n + 1). n : 2

Áp dụng công thức tính tổng S trên vào biểu thức A ta có:

\[A = \frac{1}{{\left( {2 + 1} \right) \times 2:2}} + \frac{1}{{\left( {3 + 1} \right) \times 3:2}} + ... + \frac{1}{{\left( {2020 + 1} \right) \times 2020:2}}\]

\[ = \frac{1}{{2 \times 3:2}} + \frac{1}{{3 \times 4:2}} + \frac{1}{{4 \times 5:2}} + ... + \frac{1}{{2020 \times 2021:2}}\]

\[ = \frac{2}{{2 \times 3}} + \frac{2}{{3 \times 4}} + \frac{2}{{4 \times 5}} + ... + \frac{2}{{2020 \times 2021}}\]

\[ = 2 \times \left( {\frac{1}{{2 \times 3}} + \frac{1}{{3 \times 4}} + \frac{1}{{4 \times 5}} + ... + \frac{1}{{2020 \times 2021}}} \right)\]

\[ = 2 \times \left( {\frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{4} - \frac{1}{5} + ... + \frac{1}{{2020}} - \frac{1}{{2021}}} \right)\]

\[ = 2 \cdot \left( {\frac{1}{2} - \frac{1}{{2021}}} \right)\]

\[ = 1 - \frac{2}{{2021}}\]

\[ = \frac{{2021 - 2}}{{2021}}\]

\[ = \frac{{2019}}{{2021}}\]

\[\frac{1}{{1 \cdot 2}} + \frac{1}{{2 \cdot 3}} + \frac{1}{{3 \cdot 4}} + \frac{1}{{4 \cdot 5}} + ... + \frac{1}{{n \cdot \left( {n + 1} \right)}}\]

\[ = \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + ... + \frac{1}{n} - \frac{1}{{n + 1}}\]

\[ = 1 - \frac{1}{{n + 1}} = \frac{{n + 1 - 1}}{{n + 1}} = \frac{n}{{n + 1}}\]

\[\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1.\]

\[ - \frac{1}{2} \cdot \left( {1 - 3x} \right) - \frac{1}{2} = 3x - 1\]

\[ - \left( {1 - 3x} \right) - 1 = 2\left( {3x - 1} \right)\]

\[ - 1 + 3x - 1 = 6x - 2\]

3x ‒ 6x = ‒2 + 1 + 1

3x = 0

x = 0

Vậy x = 0.

\[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + \frac{1}{{32}} + \frac{1}{{64}} + \frac{1}{{128}}\]

\[ = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{4} + \frac{1}{4} - \frac{1}{8} + \frac{1}{8} - \frac{1}{{16}} + \frac{1}{{16}} - \frac{1}{{32}} + \frac{1}{{32}} - \frac{1}{{64}} + \frac{1}{{64}} - \frac{1}{{128}}\]

\[ = 1 - \frac{1}{{128}} = \frac{{127}}{{128}}.\]

\[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + \frac{1}{{32}}\]

\[ = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{4} + \frac{1}{4} - \frac{1}{8} + \frac{1}{8} - \frac{1}{{16}} + \frac{1}{{16}} - \frac{1}{{32}}\]

\[ = 1 - \frac{1}{{32}} = \frac{{31}}{{32}}.\]

Đặt \[A = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + ... + \frac{1}{{2048}}\]

\[2A = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + ... + \frac{1}{{1024}}\]

\[2A - A = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... + \frac{1}{{1024}} - \frac{1}{2} - \frac{1}{4} - \frac{1}{8} - \frac{1}{{16}} - ... - \frac{1}{{2048}}\]

\[A = 1 - \frac{1}{{2048}} = \frac{{2047}}{{2048}}\]

\[\frac{1}{2} + \frac{1}{6} + \frac{1}{{12}} + \frac{1}{{20}} + \frac{1}{{30}} + \frac{1}{{42}}\]

Ta nhận thấy:

\[\frac{1}{2} = 1 - \frac{1}{2};\frac{1}{6} = \frac{1}{2} - \frac{1}{3};\frac{1}{{12}} = \frac{1}{3} - \frac{1}{4};\frac{1}{{20}} = \frac{1}{4} - \frac{1}{5};\frac{1}{{30}} = \frac{1}{5} - \frac{1}{6};\frac{1}{{42}} = \frac{1}{6} - \frac{1}{7}\]

Ta có:

\[\frac{1}{2} + \frac{1}{6} + \frac{1}{{12}} + \frac{1}{{20}} + \frac{1}{{30}} + \frac{1}{{42}}\]

\[ = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \frac{1}{5} - \frac{1}{6} + \frac{1}{6} - \frac{1}{7}\]

\[ = 1 - \frac{1}{7} = \frac{6}{7}.\]

\[1! + 2! + ... + 10! = 55!\]

\[P = \left( {\frac{1}{{{2^2}}} - 1} \right)\left( {\frac{1}{{{3^2}}} - 1} \right)\left( {\frac{1}{{{4^2}}} - 1} \right) \cdot \cdot \cdot \left( {\frac{1}{{{{2021}^2}}} - 1} \right)\]

\[ = \frac{{1 - {2^2}}}{{{2^2}}} \cdot \frac{{1 - {3^2}}}{{{3^2}}} \cdot \frac{{1 - {4^2}}}{{{4^2}}} \cdot \cdot \cdot \frac{{1 - {{2021}^2}}}{{{{2021}^2}}}\]

\[ = \frac{{\left( {1 - 2} \right)\left( {1 + 2} \right)}}{{{2^2}}} \cdot \frac{{\left( {1 - 3} \right)\left( {1 + 3} \right)}}{{{3^2}}} \cdot \cdot \cdot \frac{{\left( {1 - 2021} \right)\left( {1 + 2021} \right)}}{{{{2021}^2}}}\]

\[ = \frac{{\left( {1 \cdot 2 \cdot 3 \cdot \cdot \cdot 2020} \right) \cdot \left( {3 \cdot 4 \cdot 5 \cdot \cdot \cdot 2022} \right)}}{{\left( {2 \cdot 3 \cdot 4 \cdot \cdot \cdot 2021} \right)\left( {2 \cdot 3 \cdot 4 \cdot \cdot \cdot 2021} \right)}}\]

\[ = \frac{{1 \cdot 2 \cdot 3 \cdot \cdot \cdot 2020}}{{2 \cdot 3 \cdot 4 \cdot \cdot \cdot 2021}} \cdot \frac{{3 \cdot 4 \cdot 5 \cdot \cdot \cdot 2022}}{{2 \cdot 3 \cdot 4 \cdot \cdot \cdot 2021}}\]

\[ = \frac{1}{{2021}} \cdot \frac{{2022}}{2} = \frac{{1011}}{{2021}} > \frac{{1011}}{{2022}} = \frac{1}{2}\]

\[P > \frac{1}{2}\]

Đặt A = 1 + 2 + 2² + 2³ + ... + 2⁹⁹ + 2¹⁰⁰

Suy ra 2A = 2 + 2² + 2³ + ... + 2⁹⁹ + 2¹⁰¹

2A ‒ A = 2 + 2² + 2³ + ... + 2⁹⁹ + 2¹⁰¹ ‒ (1 + 2 + 2² + 2³ + ... + 2⁹⁹ + 2¹⁰⁰)

= 2¹⁰¹ ‒ 1.

Vậy 1 + 2 + 2² + 2³ + ... + 2⁹⁹ + 2¹⁰⁰ = 2¹⁰¹ ‒ 1.

S = 1² + 2² + 3² + ... + n²

= 1.2 ‒ 1 + 2.3 ‒ 2 + 3.4 ‒ 3 + ... + n(n + 1) ‒ n

= [1.2 + 2.3 + 3.4 + ... + n(n + 1)] ‒ (1 + 2 + 3 + ... + n)

Theo dạng tổng quát:

\[1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + ... + n\left( {n + 1} \right) = \frac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{3}\]

\[ \Rightarrow S = \frac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{3} - \frac{{n\left( {n + 1} \right)}}{2}\]

\[ = \frac{{2n\left( {n + 1} \right)\left( {n + 2} \right)}}{6} - \frac{{3n\left( {n + 1} \right)}}{6}\]

\[ = \frac{{2n\left( {n + 1} \right)\left( {n + 2} \right) - 3n\left( {n + 1} \right)}}{6}\]

\[ = \frac{{n\left( {n + 1} \right) \cdot \left[ {2\left( {n + 2} \right) - 3} \right]}}{6}\]

\[ = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\]

Vậy \[S = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\]