1/(1+2)+1/(1+2+3)+.....+1/1+2+3+.....+2014 = ?
Những câu hỏi liên quan
A=2014+[2014:(1+2)]+[2014:(1+2+3)]+[2014:(1+2+3+4)]+...++[2014:(1+2+3+...+2013)]
25 tháng 6 2023 lúc 8:53
Hãy so sánh M và N biết:
M 1 2014 + 2 2013 + 3 2012 + ... + 2014 1
N 1 + (1 + 2) + (1 + 2 + 3) + ... + (1 + 2 + 3 + ... + 2014)
Đọc tiếp
Hãy so sánh M và N biết:
M = 1 2014 + 2 2013 + 3 2012 + ... + 2014 1
N = 1 + (1 + 2) + (1 + 2 + 3) + ... + (1 + 2 + 3 + ... + 2014)
rut gon bieu thuc
1/(1^4+1^2+1)+2/(2^4+2^2+1)+3/(3^4+3^2+1)+...+2014/(2014^4+2014^2+2014)=...
{giup minh vs}
Cho:P =(1-1/1+2)(1-1/1+2+3)(1-1/1+2+3+4)...(1-1/1+2+3+...+2014). Tính: 2014/2016P
\(P=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+3+...+2014}\right)\)
\(P=\frac{\left(1+2\right).2:2-1}{\left(1+2\right).2:2}.\frac{\left(1+3\right).3:2-1}{\left(1+3\right).3:2}.\frac{\left(1+4\right).4:2-1}{\left(1+4\right).4:2}...\frac{\left(1+2014\right).2014:2-1}{\left(1+2014\right).2014:2}\)
\(P=\frac{2}{2.3:2}.\frac{5}{3.4:2}.\frac{9}{4.5:2}...\frac{2029104}{2014.2015:2}\)
\(P=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{2013.2016}{2014.2015}\)
\(P=\frac{1.2.3...2013}{2.3.4...2014}.\frac{4.5.6...2016}{3.4.5...2015}\)
\(P=\frac{1}{2014}.\frac{2016}{3}=\frac{1}{2014}.672=\frac{336}{1007}\)
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Bình luận (0)
1) 1/2 + 1/3 + 1/4 + ... + 1/2013 + 1/2014
2) 2014 + 2013/2 + 2012/3 + 2011/4 + ... + 2/2013 + 1/2014
frac{frac{1}{2}+frac{1}{3}+......+frac{1}{2013}}{frac{2012}{1}+frac{2012}{2}+frac{2011}{3}+...+frac{1}{2013}}frac{frac{1}{2}+frac{1}{3}+..+frac{1}{2013}}{frac{2012}{1}+2+frac{2012}{2}+1+frac{2011}{3}+1+...+frac{1}{2013}+1-2014}frac{frac{1}{2}+frac{1}{3}+...+frac{1}{2013}}{frac{2014}{1}+frac{2014}{2}+...+frac{2014}{2013}-2014}frac{frac{1}{2}+frac{1}{3}+...+frac{1}{2013}}{2014left(1+frac{1}{2}+frac{1}{3}+...+frac{1}{2013}-1right)}frac{1}{2014}
Đọc tiếp
\(\frac{\frac{1}{2}+\frac{1}{3}+......+\frac{1}{2013}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2013}}{\frac{2012}{1}+2+\frac{2012}{2}+1+\frac{2011}{3}+1+...+\frac{1}{2013}+1-2014}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{\frac{2014}{1}+\frac{2014}{2}+...+\frac{2014}{2013}-2014}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{2014\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}-1\right)}\)
=\(\frac{1}{2014}\)
Rút gọn biểu thức: \(\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+1}+\frac{3}{3^4+3^2+1}+...+\frac{2014}{2014^4+2014^2+1}\)
P= (1-1/1+2)(1-1/1+2+3)....(1-1/1+2+3..+2014)Tìm 2014/2016*P=
\(y=\frac{1}{1^4+1^2+1}+\frac{1}{2^4+2^2+1}+\frac{1}{3^4+3^2+1}+............+\frac{2014}{2014^4+2014^2+1}\)
P= (1-1/1+2)(1-1/1+2+3)....(1-1/1+2+3..+2014)Tìm 2014/2016*P=