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H2
0
NV
1
22 tháng 3 2021
Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
NV
Cho A = 1/2 + 1/3 + 1/4 + ... + 1/2017 B = 1/2015 + 2/2014 +3/2013 + ...+ 2015/2 + 2016/1 Tính B : A
1
22 tháng 3 2021
Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
NK
2
24 tháng 7 2016
Câu hỏi của Vũ Lê Ngọc Liên - Toán lớp 6 - Học toán với OnlineMath đây có câu giống nè :)
24 tháng 7 2016
\(A=\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+2015}\right)\)
\(A=\frac{2}{\left(1+2\right).2:2}.\frac{5}{\left(1+3\right).3:2}...\frac{\left(1+2015\right).2015:2-1}{\left(1+2015\right).2015:2}\)
\(A=\frac{2}{2.3:2}.\frac{5}{3.4:2}...\frac{2016.2015:2-1}{2015.2016:2}\)
\(A=\frac{4}{2.3}.\frac{10}{3.4}.\frac{\left(1008.2015-1\right).2}{2015.2016}\)
\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}...\frac{2014.2017}{2015.2016}\)
\(A=\frac{1.2...2014}{2.3...2015}.\frac{4.5...2017}{3.4...2016}\)
\(A=\frac{1}{2015}.\frac{2017}{3}=\frac{2017}{6045}\)
LT
13 tháng 7 2017
Xin lỗi nhé mình mới học lớp 6 ko biết hnhieeuf về bài lớp 7 lên mình chỉ làm được mỗi câu a thôi, nhớ tích cho mk nhé
a)
A= \(5^2+10^2+15^2+...+2015^2\)
\(A=\left(5.1\right)^2+\left(5.2\right)^2+\left(5.3\right)^2+...+\left(5.403\right)^2\)
\(A=5^2.1^2+5^2.2^2+5^2.3^2+...+5^2.403^2\)
\(A=5^2.\left(1^2+2^2+3^2+...+403^2\right)\)
\(A=25.\left[1.\left(2-1\right)+2.\left(3-1\right)+3.\left(4-1\right)+...+403.\left(404-1\right)\right]\)
\(A=25.\left[\left(1.2+2.3+3.4+...+403.404\right)-\left(1+2+3+...+403\right)\right]\)
Gọi :\(B=1.2+2.3+3.4+...+403.404\)
\(3B=1.2.3+2.3.3+3.4.3+...+403.404.3\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+403.404.\left(405-402\right)\)
\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+403.404.405-402.403.404\)
\(=403.404.405\)
\(=65938860\)
Gọi \(C=1+2+3+...+403\) (403 số hạng)
\(=\frac{\left(403+1\right).403}{2}\)
\(=\frac{162812}{2}\)
\(=81406\)
Suy ra \(A=25.\left(B-C\right)\)
\(=25.\left(65938860-81406\right)\)
\(=25.65857454\)
\(=1646436350\)
VD
0