Tính
(\(\frac{1}{2^2}\)-1)(\(\frac{1}{3^2}\)-1)......(\(\frac{1}{2018^2}\)-1)
K
Khách
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Những câu hỏi liên quan
27 tháng 9 2020
\(B=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+...+\frac{1}{2018}.\frac{\left(1+2018\right).2018}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+...+\frac{2019}{2}=1+\frac{3+4+...+2019}{2}=1+\frac{\left(3+2019\right)2017}{2}=2039188\)
T
20 tháng 3 2018
\(\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2017^2}+\frac{1}{2018^2}}\)\)
Với n thuộc N*, ta có:
\(\(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{1+\frac{1}{n^2}+\frac{2\left(n+1-n-1\right)}{n\left(n+1\right)}}\)\)
\(\(=\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}+2.1.\frac{1}{n}-2.1.\frac{1}{n+1}-2.\frac{1}{n}.\frac{1}{\left(n+1\right)}}\)\)
\(\(=\sqrt{\left(1+\frac{1}{n}-\frac{1}{n-1}\right)^2}=1+\frac{1}{n}-\frac{1}{n-1}\)\). Áp dụng vô bài, ta có:
\(\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+....+\sqrt{1+\frac{1}{2017^2}+\frac{1}{2018^2}}\)\)
\(\(=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2017}-\frac{1}{2018}\)\)
\(\(=2016+\frac{1}{2}-\frac{1}{2018}=2016\frac{504}{1009}\)\)
P/s: Lại là thằng quỷ Thắng
CH
Cô Hoàng Huyền
Admin
VIP
28 tháng 6 2017
Xét số hạng tổng quát
\(1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}=1^2+\left(\frac{1}{k}\right)^2+\left(\frac{1}{k+1}\right)^2+2.1.\frac{1}{k}-2.\left(\frac{1}{k}.\frac{1}{k+1}\right)-2.1.\frac{1}{k+1}\)
\(=\left(1+\frac{1}{k}-\frac{1}{k+1}\right)^2\)
( Vì \(\frac{1}{k}-\frac{1}{k\left(k+1\right)}-\frac{1}{k+1}=\frac{k+1-1-k}{k\left(k+1\right)}=0\) )
Vậy thì \(\sqrt{1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}}=1+\frac{1}{k}-\frac{1}{k+1}\)
Vậy \(A=\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2017^2}+\frac{1}{2018^2}}\)
\(=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2017}-\frac{1}{2018}\)
\(=2016+\frac{1}{2}-\frac{1}{2018}=2016\frac{504}{1009}\)
3 tháng 6 2020
ta có B= 1/2018+2/2017+3/2016+...+2017/2+2018/1
=> B=1+1+1+..+1( 2018 số hạng 1)+ 1/2018+..+2017/2
=> B= (1+1/2018)+(1+2/2017)+(1+3/2016)+...+(1+2017/2)+ 2019/2019
=> B= 2019 *(1/2+1/3+...+1/2019)
=> A/B= (1/2+1/3+...+1/2019)/2019*(1/2+1/3+..+1/2019)
=> A/B= 1/2019
S
12 tháng 7 2018
a, \(M=\frac{3}{2}\cdot\frac{4}{3}\cdot\cdot\cdot\cdot\frac{2018}{2017}\cdot\frac{2019}{2018}=\frac{3.4...2019}{2.3...2018}=\frac{2019}{2}\)
b, c cùng 1 câu phải k
ta có: \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=1+\frac{1}{2}+...+\frac{1}{2018}-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}=B\)
\(\Rightarrow\frac{A}{B}=1\Rightarrow\left(\frac{A}{B}\right)^{2018}=1^{2018}=1\)
LD
15 tháng 7 2018
A,\(M=\frac{3}{2}\cdot\frac{4}{3}....\frac{2018}{2017}\cdot\frac{2019}{2018}=\frac{4\cdot3...2019}{2\cdot3...2018}=\frac{2019}{2}\)
NHA
HỌC TỐT