Tinh:
S=2015+\(\frac{2015}{1+2}\)+\(\frac{2015}{1+2+3}\)+...+\(\frac{2015}{1+2+3+...+2016}\)
Những câu hỏi liên quan
Tính nhanh : \(\frac{2017+\frac{1}{2016}+\frac{2}{2015}+\frac{3}{2014}+...+\frac{2015}{2}+\frac{2016}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}}\)
RGBT:
E=\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{2015\sqrt{2014}+2014\sqrt{2015}}+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)
Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thế vô bài toán được
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)
\(=1-\frac{1}{\sqrt{2016}}\)
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Tính tổng S = \(2015+\frac{2015}{1+2}+\frac{2015}{1+2+3}+...+\frac{2015}{1+2+3+...+2016}\)
Ta có :
\(S=2015+\frac{2015}{1+2}+\frac{2015}{1+2+3}+...+\frac{2015}{1+2+3+..+2016}\)
\(=2015.\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+..+2016}\right)\)
\(=2015.\left(1+\frac{1}{\frac{\left(2+1\right).2}{2}}+\frac{1}{\frac{\left(3+1\right).3}{2}}+...+\frac{1}{\frac{\left(2016+1\right).2016}{2}}\right)\)
\(=2015.\left(\frac{2}{2}+\frac{2}{2.\left(2+1\right)}+\frac{2}{3.\left(3+1\right)}+...+\frac{2}{2016.\left(2016+1\right)}\right)\)
\(=2015.2.\left(\frac{1}{2}+\frac{1}{2.\left(2+1\right)}+\frac{1}{3.\left(3+1\right)}+...+\frac{1}{2016.\left(2016+1\right)}\right)\)
\(=2015.2.\left(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\right)\)
\(=2015.2.\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)
\(=2015.2.\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{2017}\right)\)
\(=2015.2.\left(1-\frac{1}{2017}\right)\)
\(=2015.2.\frac{2016}{2017}\)
=\(\frac{2015.2.2016}{2017}\)
=\(\frac{8124480}{2017}\)
Vậy \(S=\frac{8124480}{2017}\)
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Sai vì ngoài học tập ra còn cần phải siêng năng chăm chỉ trong các lĩnh vực khác nửa như giúp đỡ mọi người ,tham gia các hoạt động rèn luyện
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Tính tổng S = \(2015+\frac{2015}{1+2}+\frac{2015}{1+2+3}+...+\frac{2015}{1+2+3+...+2016}\)
Tính tổng:
\(S=2015+\frac{2015}{1+2}+\frac{2015}{1+2+3}+....+\frac{2015}{1+2+3+...+2016}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}}{\frac{2015}{1}+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2014}+\frac{1}{2015}}\)
xét mẫu(chỗ 1/2014 sửa lại thành 2/2014)
=(1/2015+1)+(2/2014+1)+...+(2013/3+1)+(2014/2+1)+(2015/1-2014)
=2016/2015+2016/2014+...+2016/3+2016/2+1
=2016.(1/2016+1/2015+...+1/4+1/3+1/2)
=> A= 1/2016
mún dễ hỉu hơn hãy gửi tin nhắn cho mik
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\(B=\frac{\frac{2016}{1}+\frac{2015}{2}+...+\frac{1}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(B=\frac{\frac{2016}{1}+\frac{2015}{2}+...+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}=\frac{1+\frac{2015}{2}+1+...+\frac{1}{2016}+1}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)(2016 số hạng 1 ở tử số)
\(=\frac{\frac{2017}{2017}+\frac{2017}{2}+\frac{2017}{3}+....+\frac{2017}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}=\frac{2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}=2017\)
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}}{\frac{2016}{1}+\frac{2015}{2}+....+\frac{2}{2015}+\frac{1}{2016}}\)
rút gọn nha mọi người
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}}{\left(\dfrac{2015}{2}+1\right)+...+\left(\dfrac{2}{2015}+1\right)+\left(\dfrac{1}{2016}+1\right)+1}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}}{\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}=\dfrac{1}{2017}\)
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Tìm x biết:
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x+2015=\frac{2016}{1}+\frac{2017}{2}+...+\frac{4029}{2014}+\frac{4030}{2015}\)
26 tháng 4 2015 lúc 13:54
Tìm x, biết:
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x+2015=\frac{2016}{1}+\frac{2017}{2}+...+\frac{4030}{2015}\).
\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=\left(\frac{2016}{1}-1\right)+\left(\frac{2017}{2}-1\right)+...+\left(\frac{4030}{2015}-1\right)\)
\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=\frac{2015}{1}+\frac{2015}{2}+...+\frac{2015}{2015}\)
\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=2015.\left(1+\frac{1}{2}+...+\frac{1}{2015}\right)\)
=> x = 2015
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