Tính:
S=\(\dfrac{1+2+2^2+2^3+..............+2^{2016}}{1-2^{2017}}\)
Những câu hỏi liên quan
So sánh \(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\) và \(B=2018\)
\(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\left(\dfrac{1}{2017}+1\right)+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{3}{2015}+1\right)+...+\left(\dfrac{2016}{2}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\dfrac{2018}{2017}+\dfrac{2018}{2016}+\dfrac{2018}{2015}+...+\dfrac{2018}{2}+\dfrac{2018}{2018}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}=2018\)
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Tính tổng:
\(S=\dfrac{1}{2\sqrt{1}+\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{2017\sqrt{2016}+2016\sqrt{2017}}\)
Lời giải:
Xét số hạng tổng quát:
$\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n(n+1)}(\sqrt{n}+\sqrt{n+1})}$
$=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}$
$=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}$
Do đó:
$S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}$
$=1-\frac{1}{\sqrt{2017}}$
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CMR: S = \(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+\dfrac{4}{4^4}+...+\dfrac{2016}{4^{2016}}+\dfrac{2017}{4^{2017}}\)< \(\dfrac{1}{2}\)
Cho \(T=\dfrac{2}{2^1}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{2017}{2^{2016}}+\dfrac{2018}{2^{2017}}\) . So sánh T và 3
Tính \(S=\dfrac{2016}{\sqrt{1+\dfrac{1}{2002^2}+\dfrac{1}{2001^2}}+\sqrt{1+\dfrac{1}{2001^2}+\dfrac{1}{2002^2}}+...+\sqrt{1+\dfrac{1}{2016^2}+\dfrac{1}{2017^2}}}\)
Cho S=\(\dfrac{2}{2^1}\)+\(\dfrac{3}{2^2}\)+\(\dfrac{4}{2^3}\)+...+\(\dfrac{n+1}{2^n}\)+...+\(\dfrac{2017}{2^{2016}}\)
So sánh S với 3.
\(S=\dfrac{2}{2^1}+\dfrac{3}{2^2}+...+\dfrac{2017}{2^{2016}}\)
\(\Rightarrow2S=2+\dfrac{3}{2^1}+\dfrac{4}{2^2}+...+\dfrac{2017}{2^{2015}}\)
\(\Rightarrow2S-S=\left(2+\dfrac{3}{2^1}+\dfrac{4}{2^2}+...+\dfrac{2017}{2^{2015}}\right)-\left(\dfrac{2}{2^1}+\dfrac{3}{2^2}+...+\dfrac{2017}{2^{2016}}\right)\)
\(\Leftrightarrow S=2+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2015}}-\dfrac{2017}{2^{2016}}\)
Tới đây thì đơn giản rồi nhé
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So Sánh : \(\sqrt{2017^2-1}-\sqrt{2016^2-1}\) và \(\dfrac{2.2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}\)
\(\sqrt{2017^2-1}-\sqrt{2016^2-1}=\dfrac{2017^2-1-2016^2+1}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}=\dfrac{\left(2017-2016\right)\left(2017+2016\right)}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}=\dfrac{1+2.2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}>\dfrac{2.2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}\)
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Ta có \(\left(\sqrt{2017^2-1}-\sqrt{2016^2-1}\right)\left(\sqrt{2017^2-1}+\sqrt{2016^2-1}\right)=2017^2-1+2016^2-1=2017^2-2016^2=\left(2017-2016\right)\left(2017+2016\right)=2017+2016< 2016+2016=2.2016\Rightarrow\left(\sqrt{2017^2-1}-\sqrt{2016^2-1}\right)\left(\sqrt{2017^2-1}+\sqrt{2016^2-1}\right)< 2.2016\Rightarrow\sqrt{2017^2-1}-\sqrt{2016^2-1}< \dfrac{2.2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}\)
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mình bị lộn 2017+2016>2.2016\(\Rightarrow\sqrt{2017^2-1}-\sqrt{2016^2-1}< \dfrac{2.2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}\)
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cho tổng T= \(\dfrac{2}{2^1}+\dfrac{3}{2^2}+\dfrac{4}{2^3}\) +...+\(\dfrac{2016}{2^{2015}}+\dfrac{2017}{2^{2016}}\)
so sánh T với 3
Bạn Phong Thần trả lời hay quá.
uk, cái bạn tên Phong Thần công nhận giỏi thật nha
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Giải típ hộ mik nha !
\(\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{n.\left(n+1\right)}=\dfrac{2016}{2017}=\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{n+1}=\dfrac{2016}{2017}=\dfrac{n+1-2}{2.\left(n+1\right)}=\dfrac{2016}{2017}=\dfrac{n-1}{2.\left(n+1\right)}=\dfrac{2016}{2017}=2017.\left(n-1\right)=2016.2\left(n+1\right)=...\)