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: $\frac{k}{\left(n-k\right).n}=\frac{n-\left(n-k\right)}{\left(n-k\right).n}=\frac{n}{\left(n-k\right).n}-\frac{n-k}{\left(n-k\right).n}=\frac{1}{n-k}-\frac{1}{n}$

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

b) $B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\dots +\frac{3}{2016.2019}$

c) $C=\frac{2}{1.7}+\frac{2}{7.13}+\frac{2}{13.19}+\dots +\frac{2}{2013.2019}$                                 

d) $D=\frac{7}{1.9}+\frac{7}{9.17}+\frac{7}{17.25}+\dots +\frac{7}{2011.2019}$

e) $E=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+\dots +\frac{3^2}{2017.2020}$                           

f) $F=\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}+\dots +\frac{1}{\mathrm{18.19.20}}$

Bài 2:

Tính các tổng sau:

a) $A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\dots +\frac{1}{2^{2020}}$                                                                                      

b) $B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\dots +\frac{1}{2048}$

Bài 3:

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

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

Hướng dẫn giải chi tiết

Bài 1:

Tính:

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

b) b) $B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\dots +\frac{3}{2016.2019}$

c) $C=\frac{2}{1.7}+\frac{2}{7.13}+\frac{2}{13.19}+\dots +\frac{2}{2013.2019}$                                 

d) $D=\frac{7}{1.9}+\frac{7}{9.17}+\frac{7}{17.25}+\dots +\frac{7}{2011.2019}$

e) $E=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+\dots +\frac{3^2}{2017.2020}$                           

f) $F=\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}+\dots +\frac{1}{\mathrm{18.19.20}}$

Phương pháp

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

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

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

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

$\begin{array}{c}=2017:\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\right)\\ =2017:\left(1-\frac{1}{2018}\right)\\ =2017:\frac{2017}{2018}\\ =2017.\frac{2018}{2017}=2018.\end{array}$

Vậy $x=\frac{-2}{3}$

b) $B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\dots +\frac{3}{2016.2019}$

        $\begin{array}{c}=\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+\dots +\frac{2019-2016}{2016.2019}\\ =\frac{5}{2.5}-\frac{2}{2.5}+\frac{8}{5.8}-\frac{5}{5.8}+\frac{11}{8.11}-\frac{8}{8.11}+\dots +\frac{2019}{2016.2019}-\frac{2016}{2016.2019}\\ =\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\dots +\frac{1}{2016}-\frac{1}{2019}\\ =\frac{1}{2}-\frac{1}{2019}\\ =\frac{2019-2}{2.2019}\\ =\frac{2017}{4038}.\end{array}$

c) $C=\frac{2}{1.7}+\frac{2}{7.13}+\frac{2}{13.19}+\dots +\frac{2}{2013.2019}$

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

$\begin{array}{c}\Rightarrow 3C=3\cdot \left(\frac{2}{1.7}+\frac{2}{7.13}+\frac{2}{13.19}+\dots +\frac{2}{2013.2019}\right)\\ =\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+\dots +\frac{6}{2013.2019}\end{array}$

$\begin{array}{c}=\left(\frac{1}{1}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{13}\right)+\left(\frac{1}{13}-\frac{1}{19}\right)+\dots +\left(\frac{1}{2013}-\frac{1}{2019}\right)\\ =\frac{1}{1}-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+\dots +\frac{1}{2013}-\frac{1}{2019}\\ =1-\frac{1}{2019}=\frac{2018}{2019}\end{array}$

$\Rightarrow 3C=\frac{2018}{2019}\Rightarrow C=\frac{2018}{2019}:3=\frac{2018}{2019}\cdot \frac{1}{3}=\frac{2018}{6057}$

d) $D=\frac{7}{1.9}+\frac{7}{9.17}+\frac{7}{17.25}+\dots +\frac{7}{2011.2019}$

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

Vậy $D=\frac{7063}{8076}.$

e) $E=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+\dots +\frac{3^2}{2017.2020}$

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

Vậy $E=\frac{6057}{2020}\cdot$

f) $F=\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}+\dots +\frac{1}{\mathrm{18.19.20}}$

Ta xét:

$\frac{2}{\mathrm{1.2.3}}=\frac{3-1}{\mathrm{1.2.3}}=\frac{3}{\mathrm{1.2.3}}-\frac{1}{\mathrm{1.2.3}}=\frac{1}{1.2}-\frac{1}{2.3}$

$\frac{2}{\mathrm{2.3.4}}=\frac{4-2}{\mathrm{2.3.4}}=\frac{4}{\mathrm{2.3.4}}-\frac{2}{\mathrm{2.3.4}}=\frac{1}{2.3}-\frac{1}{3.4}$

$........$

$\frac{2}{\mathrm{18.19.20}}=\frac{20-18}{\mathrm{18.19.20}}$$=\frac{20}{\mathrm{18.19.20}}-\frac{18}{\mathrm{18.19.20}}$$=\frac{1}{18.19}-\frac{1}{19.20}$

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

$\Rightarrow F=\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}+\dots +\frac{1}{\mathrm{18.19.20}}$

$\Rightarrow 2F=\frac{2}{\mathrm{1.2.3}}+\frac{2}{\mathrm{2.3.4}}+\frac{2}{\mathrm{3.4.5}}+\dots +\frac{2}{\mathrm{18.19.20}}$

     $=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+\dots +\frac{1}{18.19}-\frac{1}{19.20}$

     $=\frac{1}{1.2}-\frac{1}{19.20}=\frac{19.10-1}{19.20}=\frac{190-1}{380}=\frac{189}{380}$

$\Rightarrow F=\frac{189}{380}:2=\frac{189}{380}\cdot \frac{1}{2}=\frac{189}{760}$

Vậy $F=\frac{189}{760}\cdot$

Bài 2:

Tính các tổng sau:

a) $A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\dots +\frac{1}{2^{2020}}$                                                                                      

b) $B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\dots +\frac{1}{2048}$

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=\frac{1}{a^1}+\frac{1}{a^2}+\frac{1}{a^3}+\dots +\frac{1}{a^n}$$\Rightarrow A.a=a\left(\frac{1}{a^1}+\frac{1}{a^2}+\frac{1}{a^3}+\dots +\frac{1}{a^n}\right)$$=1+\frac{1}{a}+\dots +\frac{1}{a^{n-1}}$

Lời giải

a) $A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\dots +\frac{1}{2^{2020}}$

$\Rightarrow 2A=2\cdot \left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\dots +\frac{1}{2^{2020}}\right)$

$\Rightarrow 2A=2\cdot \frac{1}{2}+2\cdot \frac{1}{2^2}+2\cdot \frac{1}{2^3}+2\cdot \frac{1}{2^4}+\dots +2\cdot \frac{1}{2^{2020}}$

$\Rightarrow 2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\dots +\frac{1}{2^{2019}}$

$A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\dots +\frac{1}{2^{2020}}$

$\Rightarrow 2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\dots +\frac{1}{2^{2019}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\dots +\frac{1}{2^{2020}}\right)$

$\Rightarrow A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\dots +\frac{1}{2^{2019}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}-\dots -\frac{1}{2^{2020}}$

$\Rightarrow A=1-\frac{1}{2^{2020}}=\frac{2^{2020}-1}{2^{2020}}$

Vậy $A=\frac{2^{2020}-1}{2^{2020}}$.

b) $B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\dots +\frac{1}{2048}$ $=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\dots +\frac{1}{2^{11}}$

$\Rightarrow 2B=2\cdot \left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\dots +\frac{1}{2^{11}}\right)$

$\Rightarrow 2B=2.1+2\cdot \frac{1}{2}+2\cdot \frac{1}{2^2}+2\cdot \frac{1}{2^3}+2\cdot \frac{1}{2^4}+\dots +2\cdot \frac{1}{2^{11}}$

$\Rightarrow 2B=2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\dots +\frac{1}{2^{10}}$

$\Rightarrow 2B=3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\dots +\frac{1}{2^{10}}$

$B=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\dots +\frac{1}{2^{11}}$

$\Rightarrow 2B-B=2-\frac{1}{2^{11}}$$\Rightarrow B=\frac{2^{12}-1}{2^{11}}$;

Vậy $B=\frac{2^{12}-1}{2^{11}}\cdot$

Bài 3:

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

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

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: $1+2+\dots +n=\frac{n+1}{2}\cdot n=\frac{n.\left(n+1\right)}{2}$

Lời giải

a) $A=\frac{1+\left(1+2\right)+\left(1+2+3\right)+\dots +\left(1+2+3+\dots +2020\right)}{1.2020+2.2019+3.2018+\dots +2020.1}$

Ta có:

$\begin{array}{c}A=\frac{1+\left(1+2\right)+\left(1+2+3\right)+\dots +\left(1+2+3+\dots +2020\right)}{1.2020+2.2019+3.2018+\dots +2020.1}\\ =\frac{1+1+2+1+2+3+\dots +1+2+3+\dots 2020}{1.2020+2.2019+3.2018+\dots +2020.1}\\ =\frac{\left(1+1+1+\dots +1\right)+\left(2+2+\dots 2\right)+\left(3+3+3+3\dots \right)+\dots +2020}{1.2020+2.2019+3.2018+\dots +2020.1}\end{array}$

$\begin{array}{c}=\frac{1.2020+2.2019+3.2018+\dots +2020.1}{1.2020+2.2019+3.2018+\dots +2020.1}\\ =1\end{array}$

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

Với biểu thức $B$, xét tử số ta có:

$1.2020+2.2019+3.2018+\dots +2020.1$

$=1+\left(1+2\right)+\left(1+2+3\right)+\dots +\left(1+2+3+\dots +2020\right)$

$=\frac{0+1}{2}\cdot 2+\frac{1+2}{2}\cdot 2+\frac{1+3}{2}\cdot 3+\dots +\frac{1+2020}{2}\cdot 2020$

$=\frac{1}{2}\cdot 2+\frac{3}{2}\cdot 2+\frac{4}{2}\cdot 3+\dots +\frac{2021}{2}\cdot 2020$

$=\frac{1.2}{2}+\frac{2.3}{2}+\frac{3.4}{2}+\dots +\frac{2020.2021}{2}$

$=\frac{1}{2}\cdot \left(1.2+2.3+3.4+\dots +2020.2021\right)$

$\begin{array}{c}\Rightarrow B=\frac{1.2020+2.2019+3.2018+\dots +2020.1}{1.2+2.3+3.4+\dots +2020.2021}\\ =\frac{\frac{1}{2}\cdot \left(1.2+2.3+3.4+\dots +2020.2021\right)}{1.2+2.3+3.4+\dots +2020.2021}=\frac{1}{2}.\end{array}$

Vậy $B=\frac{1}{2}\cdot$