Dạng 6. Dãy phân số viết theo quy luật Chủ đề 6 Ôn hè Toán 6

Lý thuyết

Phát hiện quy luật của dãy số

Dạng tổng quát: \(\dfrac{k}{{\left( {n - k} \right).n}} = \dfrac{{n - \left( {n - k} \right)}}{{\left( {n - k} \right).n}} = \dfrac{n}{{\left( {n - k} \right).n}} - \dfrac{{n - k}}{{\left( {n - k} \right).n}} = \dfrac{1}{{n - k}} - \dfrac{1}{n}\)

Áp dụng phương pháp khử liên tiếp: Viết mỗi số hạng thành hiệu của hai số sao cho số trừ ở nhóm trước bằng số bị trừ ở nhóm sau.

Bài tập

Bài 1:

Tính:

a) A = \(2017:\left( {\dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + ... + \dfrac{1}{{2017.2018}}} \right)\)

b) \(B = \dfrac{3}{{2.5}} + \dfrac{3}{{5.8}} + \dfrac{3}{{8.11}} +  \ldots  + \dfrac{3}{{2016.2019}}\)

c) \(C = \dfrac{2}{{1.7}} + \dfrac{2}{{7.13}} + \dfrac{2}{{13.19}} +  \ldots  + \dfrac{2}{{2013.2019}}\)                                 

d) \(D = \dfrac{7}{{1.9}} + \dfrac{7}{{9.17}} + \dfrac{7}{{17.25}} +  \ldots  + \dfrac{7}{{2011.2019}}\)

e) \(E = \dfrac{{{3^2}}}{{1.4}} + \dfrac{{{3^2}}}{{4.7}} + \dfrac{{{3^2}}}{{7.10}} +  \ldots  + \dfrac{{{3^2}}}{{2017.2020}}\)                           

f) \(F = \dfrac{1}{{1.2.3}} + \dfrac{1}{{2.3.4}} + \dfrac{1}{{3.4.5}} +  \ldots  + \dfrac{1}{{18.19.20}}\)

Bài 2:

Tính các tổng sau:

a) \(A = \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \dfrac{1}{{{2^4}}} +  \ldots  + \dfrac{1}{{{2^{2020}}}}\)                                                                                      

b) \(B = 1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{{16}} + \dfrac{1}{{32}} +  \ldots  + \dfrac{1}{{2048}}\)

Bài 3:

a) Tính tổng sau: \(A = \dfrac{{1 + \left( {1 + 2} \right) + \left( {1 + 2 + 3} \right) +  \ldots  + \left( {1 + 2 + 3 +  \ldots  + 2020} \right)}}{{1.2020 + 2.2019 + 3.2018 +  \ldots  + 2020.1}}\)

b) Chứng minh rằng biểu thức \(B\) có giá trị bằng \(\dfrac{1}{2}\) với \(B = \dfrac{{1.2020 + 2.2019 + 3.2018 +  \ldots  + 2020.1}}{{1.2 + 2.3 + 3.4 +  \ldots  + 2020.2021}}.\)

Hướng dẫn giải chi tiết

Bài 1:

Tính:

a) A = \(2017:\left( {\dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + ... + \dfrac{1}{{2017.2018}}} \right)\)

b) b) \(B = \dfrac{3}{{2.5}} + \dfrac{3}{{5.8}} + \dfrac{3}{{8.11}} +  \ldots  + \dfrac{3}{{2016.2019}}\)

c) \(C = \dfrac{2}{{1.7}} + \dfrac{2}{{7.13}} + \dfrac{2}{{13.19}} +  \ldots  + \dfrac{2}{{2013.2019}}\)                                 

d) \(D = \dfrac{7}{{1.9}} + \dfrac{7}{{9.17}} + \dfrac{7}{{17.25}} +  \ldots  + \dfrac{7}{{2011.2019}}\)

e) \(E = \dfrac{{{3^2}}}{{1.4}} + \dfrac{{{3^2}}}{{4.7}} + \dfrac{{{3^2}}}{{7.10}} +  \ldots  + \dfrac{{{3^2}}}{{2017.2020}}\)                           

f) \(F = \dfrac{1}{{1.2.3}} + \dfrac{1}{{2.3.4}} + \dfrac{1}{{3.4.5}} +  \ldots  + \dfrac{1}{{18.19.20}}\)

Phương pháp

Nhận xét: Tử số bằng hiệu của các thừa số ở mẫu.

Dạng tổng quát: \(\dfrac{k}{{\left( {n - k} \right).n}} = \dfrac{{n - \left( {n - k} \right)}}{{\left( {n - k} \right).n}} = \dfrac{n}{{\left( {n - k} \right).n}} - \dfrac{{n - k}}{{\left( {n - k} \right).n}} = \dfrac{1}{{n - k}} - \dfrac{1}{n}\)

Áp dụng phương pháp khử liên tiếp: Viết mỗi số hạng thành hiệu của hai số sao cho số trừ ở nhóm trước bằng số bị trừ ở nhóm sau.

Lời giải

\(2017:\left( {\dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + ... + \dfrac{1}{{2017.2018}}} \right)\)

\(\begin{array}{*{20}{l}}{ = 2017:\left( {1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + ... + \dfrac{1}{{2017}} - \dfrac{1}{{2018}}} \right)}\\{ = 2017:\left( {1 - \dfrac{1}{{2018}}} \right)}\\{ = 2017:\dfrac{{2017}}{{2018}}}\\{ = 2017.\dfrac{{2018}}{{2017}} = 2018.}\end{array}\)

Vậy \(x = \dfrac{{ - 2}}{3}\)

b) \(B = \dfrac{3}{{2.5}} + \dfrac{3}{{5.8}} + \dfrac{3}{{8.11}} +  \ldots  + \dfrac{3}{{2016.2019}}\)

        \(\begin{array}{l} = \dfrac{{5 - 2}}{{2.5}} + \dfrac{{8 - 5}}{{5.8}} + \dfrac{{11 - 8}}{{8.11}} +  \ldots  + \dfrac{{2019 - 2016}}{{2016.2019}}\\\, = \dfrac{5}{{2.5}} - \dfrac{2}{{2.5}} + \dfrac{8}{{5.8}} - \dfrac{5}{{5.8}} + \dfrac{{11}}{{8.11}} - \dfrac{8}{{8.11}} +  \ldots  + \dfrac{{2019}}{{2016.2019}} - \dfrac{{2016}}{{2016.2019}}\\\, = \dfrac{1}{2} - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{8} + \dfrac{1}{8} - \dfrac{1}{{11}} +  \ldots  + \dfrac{1}{{2016}} - \dfrac{1}{{2019}}\\\, = \dfrac{1}{2} - \dfrac{1}{{2019}}\\\, = \dfrac{{2019 - 2}}{{2.2019}}\\\, = \dfrac{{2017}}{{4038}}.\end{array}\)

c) \(C = \dfrac{2}{{1.7}} + \dfrac{2}{{7.13}} + \dfrac{2}{{13.19}} +  \ldots  + \dfrac{2}{{2013.2019}}\)

Xét từng phân số ta thấy: Hiệu 2 thừa số ở mẫu bằng \(6\) \( \Rightarrow \) Nhân cả 2 vế của biểu thức với \(3\).

\(\begin{array}{l} \Rightarrow 3C = 3 \cdot \left( {\dfrac{2}{{1.7}} + \dfrac{2}{{7.13}} + \dfrac{2}{{13.19}} +  \ldots  + \dfrac{2}{{2013.2019}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{6}{{1.7}} + \dfrac{6}{{7.13}} + \dfrac{6}{{13.19}} +  \ldots  + \dfrac{6}{{2013.2019}}\end{array}\)

\(\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\dfrac{1}{1} - \dfrac{1}{7}} \right) + \left( {\dfrac{1}{7} - \dfrac{1}{{13}}} \right) + \left( {\dfrac{1}{{13}} - \dfrac{1}{{19}}} \right) +  \ldots  + \left( {\dfrac{1}{{2013}} - \dfrac{1}{{2019}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{1} - \dfrac{1}{7} + \dfrac{1}{7} - \dfrac{1}{{13}} + \dfrac{1}{{13}} - \dfrac{1}{{19}} +  \ldots  + \dfrac{1}{{2013}} - \dfrac{1}{{2019}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - \dfrac{1}{{2019}} = \dfrac{{2018}}{{2019}}\end{array}\)

\( \Rightarrow 3C = \dfrac{{2018}}{{2019}} \Rightarrow C = \dfrac{{2018}}{{2019}}:3 = \dfrac{{2018}}{{2019}} \cdot \dfrac{1}{3} = \dfrac{{2018}}{{6057}}\)

d) \(D = \dfrac{7}{{1.9}} + \dfrac{7}{{9.17}} + \dfrac{7}{{17.25}} +  \ldots  + \dfrac{7}{{2011.2019}}\)

\(\begin{array}{l} \Rightarrow D = 7 \cdot \dfrac{8}{8} \cdot \left( {\dfrac{1}{{1.9}} + \dfrac{1}{{9.17}} + \dfrac{1}{{17.25}} +  \ldots  + \dfrac{1}{{2011.2019}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{7}{8} \cdot \left( {\dfrac{8}{{1.9}} + \dfrac{8}{{9.17}} + \dfrac{8}{{17.25}} +  \ldots  + \dfrac{8}{{2011.2019}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{7}{8} \cdot \left( {1 - \dfrac{1}{9} + \dfrac{1}{9} - \dfrac{1}{{17}} + \dfrac{1}{{17}} - \dfrac{1}{{25}} +  \ldots  + \dfrac{1}{{2011}} - \dfrac{1}{{2019}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{7}{8} \cdot \left( {1 - \dfrac{1}{{2019}}} \right) = \dfrac{7}{8} \cdot \left( {\dfrac{{2019}}{{2019}} - \dfrac{1}{{2019}}} \right)\\\,\,\,\,\,\,\,\,\,\,\, = \dfrac{7}{8} \cdot \dfrac{{2018}}{{2019}} = \dfrac{{7.1009}}{{4.2019}} = \dfrac{{7063}}{{8076}}\end{array}\)

Vậy \(D = \dfrac{{7063}}{{8076}}.\)

e) \(E = \dfrac{{{3^2}}}{{1.4}} + \dfrac{{{3^2}}}{{4.7}} + \dfrac{{{3^2}}}{{7.10}} +  \ldots  + \dfrac{{{3^2}}}{{2017.2020}}\)

       \(\begin{array}{l} = \dfrac{{3.3}}{{1.4}} + \dfrac{{3.3}}{{4.7}} + \dfrac{{3.3}}{{7.10}} +  \ldots  + \dfrac{{3.3}}{{2017.2020}}\\ = 3 \cdot \left( {\dfrac{3}{{1.4}} + \dfrac{3}{{4.7}} + \dfrac{3}{{7.10}} +  \ldots  + \dfrac{3}{{2017.2020}}} \right)\\ = 3 \cdot \left( {\dfrac{1}{1} - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{7} + \dfrac{1}{7} - \dfrac{1}{{10}} +  \ldots  + \dfrac{1}{{2017}} - \dfrac{1}{{2020}}} \right)\\ = 3 \cdot \left( {1 - \dfrac{1}{{2020}}} \right) = 3 \cdot \left( {\dfrac{{2020}}{{2020}} - \dfrac{1}{{2020}}} \right)\\ = 3 \cdot \dfrac{{2019}}{{2020}} = \dfrac{{6057}}{{2020}}\end{array}\)

Vậy \(E = \dfrac{{6057}}{{2020}} \cdot \)

f) \(F = \dfrac{1}{{1.2.3}} + \dfrac{1}{{2.3.4}} + \dfrac{1}{{3.4.5}} +  \ldots  + \dfrac{1}{{18.19.20}}\)

Ta xét:

\(\dfrac{2}{{1.2.3}} = \dfrac{{3 - 1}}{{1.2.3}} = \dfrac{3}{{1.2.3}} - \dfrac{1}{{1.2.3}} = \dfrac{1}{{1.2}} - \dfrac{1}{{2.3}}\)

\(\dfrac{2}{{2.3.4}} = \dfrac{{4 - 2}}{{2.3.4}} = \dfrac{4}{{2.3.4}} - \dfrac{2}{{2.3.4}} = \dfrac{1}{{2.3}} - \dfrac{1}{{3.4}}\)

\(........\)

\(\dfrac{2}{{18.19.20}} = \dfrac{{20 - 18}}{{18.19.20}}\)\( = \dfrac{{20}}{{18.19.20}} - \dfrac{{18}}{{18.19.20}}\)\( = \dfrac{1}{{18.19}} - \dfrac{1}{{19.20}}\)

Tổng quát: \(\dfrac{1}{{n.\left( {n + 1} \right)}} - \dfrac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}} = \dfrac{2}{{n\left( {n + 1} \right)\left( {n + 2} \right)}}\)

\( \Rightarrow F = \dfrac{1}{{1.2.3}} + \dfrac{1}{{2.3.4}} + \dfrac{1}{{3.4.5}} +  \ldots  + \dfrac{1}{{18.19.20}}\)

\( \Rightarrow 2F = \dfrac{2}{{1.2.3}} + \dfrac{2}{{2.3.4}} + \dfrac{2}{{3.4.5}} +  \ldots  + \dfrac{2}{{18.19.20}}\)

     \(\,\,\,\,\,\,\,\, = \dfrac{1}{{1.2}} - \dfrac{1}{{2.3}} + \dfrac{1}{{2.3}} - \dfrac{1}{{3.4}} + \dfrac{1}{{3.4}} - \dfrac{1}{{4.5}} +  \ldots  + \dfrac{1}{{18.19}} - \dfrac{1}{{19.20}}\)

     \(\,\,\,\,\,\,\,\, = \dfrac{1}{{1.2}} - \dfrac{1}{{19.20}} = \dfrac{{19.10 - 1}}{{19.20}} = \dfrac{{190 - 1}}{{380}} = \dfrac{{189}}{{380}}\)

\( \Rightarrow F = \dfrac{{189}}{{380}}:2 = \dfrac{{189}}{{380}} \cdot \dfrac{1}{2} = \dfrac{{189}}{{760}}\)

Vậy \(F = \dfrac{{189}}{{760}} \cdot \)

Bài 2:

Tính các tổng sau:

a) \(A = \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \dfrac{1}{{{2^4}}} +  \ldots  + \dfrac{1}{{{2^{2020}}}}\)                                                                                      

b) \(B = 1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{{16}} + \dfrac{1}{{32}} +  \ldots  + \dfrac{1}{{2048}}\)

Phương pháp

Xét các phân số có tử bằng nhau và có mẫu là lũy thừa tăng dần của cùng 1 cơ số thì ta nhân cả 2 vế với đúng cơ số đó. Trường hợp tổng quát:

\(A = \dfrac{1}{{{a^1}}} + \dfrac{1}{{{a^2}}} + \dfrac{1}{{{a^3}}} +  \ldots  + \dfrac{1}{{{a^n}}}\)\( \Rightarrow A.a = a\left( {\dfrac{1}{{{a^1}}} + \dfrac{1}{{{a^2}}} + \dfrac{1}{{{a^3}}} +  \ldots  + \dfrac{1}{{{a^n}}}} \right)\)\( = 1 + \dfrac{1}{a} +  \ldots  + \dfrac{1}{{{a^{n - 1}}}}\)

Lời giải

a) \(A = \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \dfrac{1}{{{2^4}}} +  \ldots  + \dfrac{1}{{{2^{2020}}}}\)

\( \Rightarrow 2A = 2 \cdot \left( {\dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \dfrac{1}{{{2^4}}} +  \ldots  + \dfrac{1}{{{2^{2020}}}}} \right)\)

\( \Rightarrow 2A = 2 \cdot \dfrac{1}{2} + 2 \cdot \dfrac{1}{{{2^2}}} + 2 \cdot \dfrac{1}{{{2^3}}} + 2 \cdot \dfrac{1}{{{2^4}}} +  \ldots  + 2 \cdot \dfrac{1}{{{2^{2020}}}}\)

\( \Rightarrow 2A = 1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} +  \ldots  + \dfrac{1}{{{2^{2019}}}}\)

\(\,\,\,\,\,\,\,\,\,A = \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \dfrac{1}{{{2^4}}} +  \ldots  + \dfrac{1}{{{2^{2020}}}}\)

\( \Rightarrow 2A - A = \left( {1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} +  \ldots  + \dfrac{1}{{{2^{2019}}}}} \right) - \left( {\dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \dfrac{1}{{{2^4}}} +  \ldots  + \dfrac{1}{{{2^{2020}}}}} \right)\)

\( \Rightarrow A = 1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} +  \ldots  + \dfrac{1}{{{2^{2019}}}} - \dfrac{1}{2} - \dfrac{1}{{{2^2}}} - \dfrac{1}{{{2^3}}} - \dfrac{1}{{{2^4}}} -  \ldots  - \dfrac{1}{{{2^{2020}}}}\)

\( \Rightarrow A = 1 - \dfrac{1}{{{2^{2020}}}} = \dfrac{{{2^{2020}} - 1}}{{{2^{2020}}}}\)

Vậy \(A = \dfrac{{{2^{2020}} - 1}}{{{2^{2020}}}}\).

b) \(B = 1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{{16}} +  \ldots  + \dfrac{1}{{2048}}\) \( = 1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \dfrac{1}{{{2^4}}} +  \ldots  + \dfrac{1}{{{2^{11}}}}\)

\( \Rightarrow 2B = 2 \cdot \left( {1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \dfrac{1}{{{2^4}}} +  \ldots  + \dfrac{1}{{{2^{11}}}}} \right)\)

\( \Rightarrow 2B = 2.1 + 2 \cdot \dfrac{1}{2} + 2 \cdot \dfrac{1}{{{2^2}}} + 2 \cdot \dfrac{1}{{{2^3}}} + 2 \cdot \dfrac{1}{{{2^4}}} +  \ldots  + 2 \cdot \dfrac{1}{{{2^{11}}}}\)

\( \Rightarrow 2B = 2 + 1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} +  \ldots  + \dfrac{1}{{{2^{10}}}}\)

\( \Rightarrow 2B = 3 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} +  \ldots  + \dfrac{1}{{{2^{10}}}}\)

\(\,\,\,\,\,\,\,\,\,B = 1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \dfrac{1}{{{2^4}}} +  \ldots  + \dfrac{1}{{{2^{11}}}}\)

\( \Rightarrow 2B - B = 2 - \dfrac{1}{{{2^{11}}}}\)\( \Rightarrow B = \dfrac{{{2^{12}} - 1}}{{{2^{11}}}}\);

Vậy \(B = \dfrac{{{2^{12}} - 1}}{{{2^{11}}}} \cdot \)

Bài 3:

a) Tính tổng sau: \(A = \dfrac{{1 + \left( {1 + 2} \right) + \left( {1 + 2 + 3} \right) +  \ldots  + \left( {1 + 2 + 3 +  \ldots  + 2020} \right)}}{{1.2020 + 2.2019 + 3.2018 +  \ldots  + 2020.1}}\)

b) Chứng minh rằng biểu thức \(B\) có giá trị bằng \(\dfrac{1}{2}\) với \(B = \dfrac{{1.2020 + 2.2019 + 3.2018 +  \ldots  + 2020.1}}{{1.2 + 2.3 + 3.4 +  \ldots  + 2020.2021}}.\)

Phương pháp

+) Áp dụng quy tắc dấu ngoặc, nhóm các hạng tử.

+) Áp dụng công thức tính tổng của 1 dãy các số tự nhiên liên tiếp: \(1 + 2 +  \ldots  + n = \dfrac{{n + 1}}{2} \cdot n = \dfrac{{n.\left( {n + 1} \right)}}{2}\)

Lời giải

a) \(A = \dfrac{{1 + \left( {1 + 2} \right) + \left( {1 + 2 + 3} \right) +  \ldots  + \left( {1 + 2 + 3 +  \ldots  + 2020} \right)}}{{1.2020 + 2.2019 + 3.2018 +  \ldots  + 2020.1}}\)

Ta có:

\(\begin{array}{l}A = \dfrac{{1 + \left( {1 + 2} \right) + \left( {1 + 2 + 3} \right) +  \ldots  + \left( {1 + 2 + 3 +  \ldots  + 2020} \right)}}{{1.2020 + 2.2019 + 3.2018 +  \ldots  + 2020.1}}\\\,\,\,\,\, = \dfrac{{1 + 1 + 2 + 1 + 2 + 3 +  \ldots  + 1 + 2 + 3 +  \ldots 2020}}{{1.2020 + 2.2019 + 3.2018 +  \ldots  + 2020.1}}\\\,\,\,\,\, = \dfrac{{\left( {1 + 1 + 1 +  \ldots  + 1} \right) + \left( {2 + 2 +  \ldots 2} \right) + \left( {3 + 3 + 3 + 3 \ldots } \right) +  \ldots  + 2020}}{{1.2020 + 2.2019 + 3.2018 +  \ldots  + 2020.1}}\end{array}\)

\(\begin{array}{l}\,\,\,\,\, = \dfrac{{1.2020 + 2.2019 + 3.2018 +  \ldots  + 2020.1}}{{1.2020 + 2.2019 + 3.2018 +  \ldots  + 2020.1}}\\\,\,\,\,\, = 1\end{array}\)

b) Chứng minh rằng biểu thức \(B\) có giá trị bằng \(\dfrac{1}{2}\) với \(B = \dfrac{{1.2020 + 2.2019 + 3.2018 +  \ldots  + 2020.1}}{{1.2 + 2.3 + 3.4 +  \ldots  + 2020.2021}}.\)

Với biểu thức \(B\), xét tử số ta có:

\(\,\,\,\,1.2020 + 2.2019 + 3.2018 +  \ldots  + 2020.1\)

\( = 1 + \left( {1 + 2} \right) + \left( {1 + 2 + 3} \right) +  \ldots  + \left( {1 + 2 + 3 +  \ldots  + 2020} \right)\)

\( = \dfrac{{0 + 1}}{2} \cdot 2 + \dfrac{{1 + 2}}{2} \cdot 2 + \dfrac{{1 + 3}}{2} \cdot 3 +  \ldots  + \dfrac{{1 + 2020}}{2} \cdot 2020\)

\( = \dfrac{1}{2} \cdot 2 + \dfrac{3}{2} \cdot 2 + \dfrac{4}{2} \cdot 3 +  \ldots  + \dfrac{{2021}}{2} \cdot 2020\)

\( = \dfrac{{1.2}}{2} + \dfrac{{2.3}}{2} + \dfrac{{3.4}}{2} +  \ldots  + \dfrac{{2020.2021}}{2}\)

\( = \dfrac{1}{2} \cdot \left( {1.2 + 2.3 + 3.4 +  \ldots  + 2020.2021} \right)\)

\(\begin{array}{l} \Rightarrow B = \dfrac{{1.2020 + 2.2019 + 3.2018 +  \ldots  + 2020.1}}{{1.2 + 2.3 + 3.4 +  \ldots  + 2020.2021}}\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{\dfrac{1}{2} \cdot \left( {1.2 + 2.3 + 3.4 +  \ldots  + 2020.2021} \right)}}{{1.2 + 2.3 + 3.4 +  \ldots  + 2020.2021}} = \dfrac{1}{2}.\end{array}\)

Vậy \(B = \dfrac{1}{2} \cdot \)