C=(1+1+1+...+1)+(1/1*3+1/2*4+1/3*5+...+1/2015*2017+1/2015*2017)

C=2015+(2/1*3+2/2*4+2/3*5+...+2/2015*2017+2/2015*2017):2

C=2015+(1-1/3+1/2-1/4+...+1/2015-1/2017+1/2015-1/2017):2

C=2015+(1+1/2-1/2016-1/2017+1/2015-1/2017)

cai nay thi ban tu tinh lay 

nho k cho minh voi nhe

Dương Thanh Minh lạc đề rùi

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+.....+\frac{1}{2016\cdot2017}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.........+\frac{1}{2016}-\frac{1}{2017}\)

\(=1-\frac{1}{2017}=\frac{2016}{2017}\)

= 1/1-1/2+1/2-1/3+1/3-............-1/2017

=1-1/2017

=2016/2017

\(B1\)

\(=\frac{1}{1}-\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{37}-\frac{1}{38}-\frac{1}{39}\)

\(=1-\frac{1}{39}\)

\(=\frac{38}{39}\)

\(B2\)

\(=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+.....+\frac{1}{99\cdot100}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+......+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{4}-\frac{1}{100}\)

\(=\frac{25}{100}-\frac{1}{100}\)

\(=\frac{24}{100}\)

\(=\frac{6}{25}\)

Bài 1 :

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\)

\(=\frac{1}{1.2}-\frac{1}{38.39}\)

\(=\frac{370}{741}\)

\(B=\frac{2.2}{1.3}.\frac{3.3}{2.4}...\frac{2015.2015}{2014.2016}\)

\(B=\frac{2.3...2015}{1.2...2014}.\frac{2.3...2015}{3.4...2016}\)

\(B=2015.\frac{1}{1008}\)

\(B=\frac{2015}{1008}\)

\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}\)

\(A=\frac{1}{4.6}+\frac{1}{10.12}+\frac{1}{18.20}+...+\frac{1}{810.812}\)

.......

~ Chúc học tốt ~ 

Ai ngang qua xin để lại 1 L - I - K - E

\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+.....+\frac{1}{27.28.29.30}\)

\(3A=3.\left(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+......+\frac{1}{27.28.29.30}\right)\)

\(3A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+..........+\frac{3}{27.28.29.30}\)

\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+........+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)

\(3A=\frac{1}{1.2.3}-\frac{1}{28.29.30}\)

\(3A=\frac{1}{6}-\frac{1}{24360}\)

\(3A=\frac{1353}{8120}\)

\(A=\frac{1353}{8120}:3\)

\(A=\frac{451}{8120}\)

a) Đặt A= \(\frac{1+2+2^2+2^3+...+2^{2009}}{1-2^{2010}}\)

Đặt S = 1 + 2 + 2² + 2³ + ... + 2²⁰⁰⁹

=> 2S = 2 + 2² + 2³ + ... + 2²⁰¹⁰

=> 2S - S = (2 + 2² + 2³ + ... + 2²⁰¹⁰) - (1 + 2 + 2² + 2³ + .. + 2²⁰⁰⁹)

=> S = 2²⁰¹⁰ - 1

=> S = - 1  - 2²⁰¹⁰

^{\(\Rightarrow A=\frac{-1-2^{2010}}{1-2^{2010}}=-1\)}

Vậy \(\frac{1+2+2^2+2^3+...+2^{2009}}{1-2^{2010}}=-1\)

b) Đặt: \(A=\frac{1}{299.297}-\frac{1}{297.295}-\frac{1}{295.293}-...-\frac{1}{3.1}\)

\(\Rightarrow-2A=-\frac{2}{299.297}+\frac{2}{297.295}+\frac{2}{295.293}+...+\frac{2}{3.1}\)

\(\Rightarrow-2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{295.297}-\frac{2}{297.299}\)

\(\Rightarrow-2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{295}-\frac{1}{297}-\frac{1}{297.299}\)

\(\Rightarrow-2A=1-\frac{1}{297}-\frac{2}{88803}\)

\(\Rightarrow-2A=\frac{296}{297}-\frac{2}{88803}=\frac{88504}{88803}-\frac{2}{88803}=\frac{88502}{88803}\)

\(\Rightarrow A=\frac{88502}{88803}:\left(-2\right)=\frac{44251}{88803}\)

Vậy \(\frac{1}{299.297}-\frac{1}{297.295}-\frac{1}{295.293}-...-\frac{1}{3.1}=\frac{44251}{88803}\)

c) Đặt \(B=\frac{12}{1.3.5}+\frac{12}{3.5.7}+\frac{12}{5.7.9}+...+\frac{12}{25.27.29}\)

\(\Rightarrow\frac{B}{3}=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{12}{25.27.29}\)

\(\Rightarrow\frac{B}{3}=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\)

\(\Rightarrow\frac{B}{3}=\frac{1}{1.3}-\frac{1}{27.29}\)

\(\Rightarrow\frac{B}{3}=\frac{1}{3}-\frac{1}{783}=\frac{261}{783}-\frac{1}{783}=\frac{260}{783}\)

\(\Rightarrow B=\frac{260}{783}.3=\frac{260}{261}\)

Vậy \(\frac{12}{1.3.5}+\frac{12}{3.5.7}+\frac{12}{5.7.9}+...+\frac{12}{25.27.29}=\frac{260}{261}\)

Duyệt mk nha!!!

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