\(ChoA=\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right).\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+3+...+2014}\right)\)
Tính A=?
Giải ra cụ thể cho mình nhé!
Những câu hỏi liên quan
\(A=\frac{\left(1-2\right).\left(1+2\right)}{2^2}.\frac{\left(1-3\right).\left(1+3\right)}{3^2}.......\frac{\left(1-2013\right).\left(1+2013\right)}{2013^2}.\frac{\left(1-2014\right).\left(1+2014\right)}{2014^2}\)
Tìm x biết
\(a,-\frac{1}{2}\left(3x-1\right)+\frac{3}{4}\left(3-2x\right)=-3\left(\frac{x}{2}-1\right)-\left(\frac{4}{5}\right)^{-1}\)
\(b,\sqrt{9\left(5x-1\right)}-\sqrt{16\cdot\left(5x-1\right)}+\sqrt{36\left(5x-1\right)}=15\)
MÌNH ĐANG CẦN GẤP GIẢI CỤ THỂ GIÚP MÌNH NHA
Tính \(M=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+4\right)+...+\frac{1}{2014}.\left(1+2+3+...+2014\right)\)
Tính A=\(\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right).\left(1-\frac{1}{1+2+3+4}\right).....\left(1-\frac{1}{1+2+3+...+2014}\right)\)
\(A=\left(\frac{1}{1+2}\right).\left(\frac{1}{1+2+3}\right).....\left(\frac{1}{1+2+3+...+2014}\right)\)
\(A=\left(\frac{1}{\frac{2.\left(2+1\right)}{2}}\right).\left(\frac{1}{\frac{3.\left(3+1\right)}{2}}\right).....\left(\frac{1}{\frac{2014.\left(2014+1\right)}{2}}\right)\)
\(A=\frac{1}{\frac{2.3}{2}}.\frac{1}{\frac{3.4}{2}}.\frac{1}{\frac{4.5}{2}}.....\frac{1}{\frac{2014.2015}{2}}\)
\(A=\frac{2}{2.3}.\frac{2}{3.4}.\frac{2}{4.5}.....\frac{2}{2014.2015}\)
Đến đây thì không tính được nữa , có thể bạn chép nhầm dấu cộng thành dấu nhân rồi.
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Nếu đổi dấu nhân thành dấu cộng, ta được:
\(A=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2014.2015}\)
\(A=2.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{2014.2015}\right)\)
\(A=2.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(A=2.\left(\frac{1}{3}-\frac{1}{2015}\right)\)
\(A=2.\frac{2012}{6045}\)
\(A=\frac{4024}{6045}\)
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Cho B=\(1+\frac{-1}{2}+\left(\frac{-1}{2}\right)^2+\left(\frac{-1}{2}\right)^3\) + \(\left(\frac{-1}{2}\right)^4+\left(\frac{-1}{2}\right)^5+........+\left(\frac{-1}{2}\right)^{98}+\left(\frac{-1}{2}\right)^{99}\)
Chứng minh B < \(\frac{2}{3}\)
MÌNH ĐANG CẦN GẤP NÊN BẠN NÀO ĐÓ GHI BÀI GIẢI RA HỘ MÌNH LUÔN NHÉ, CẢM ƠN BẠN TRƯỚC
tính
a, A=\(\frac{2\times2306}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2306}}\)
b, B=\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{2014}-1\right)\left(\frac{1}{2015}-1\right)\)
Giải rõ giùm nghen
Tính \(S=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2}+\frac{1}{3}\right)\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+...+\left(1+\frac{1}{2}+...+\frac{1}{10}\right)\)
Các bạn giúp mình nhé !~~~
Tính A = \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2014}\right)\left(1-\frac{1}{2015}\right)\left(1-\frac{1}{2016}\right)\)
Ta có :
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2016}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2015}{2016}\)
\(A=\frac{2.3.4.....2015}{2.3.4.....2015}.\frac{1}{2016}\)
\(A=\frac{1}{2016}\)
Vậy \(A=\frac{1}{2016}\)
Chúc bạn học tốt ~
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\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)..\left(1-\frac{1}{2016}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2015}{2016}\)
\(\Rightarrow A=\frac{1.2.3..2015}{2.3.4..2016}\)
\(\Rightarrow A=\frac{1}{2016}\)
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k đúng cho mk nha
Xem thêm câu trả lời
\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2013^2}-1\right).\left(\frac{1}{2014^2}-1\right)\)
\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)....\left(\frac{1}{100^2}-1\right)\)( có 2013 thừa số )
\(A=\left(-\frac{3}{2^2}\right).\left(-\frac{8}{3^2}\right).\left(-\frac{15}{4^2}\right).....\left(-\frac{\text{4056196}}{2014^2}\right)\)
\(-A=\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{4056196}{2014^2}=\frac{1.3.2.4.3.5....2013.2015}{2.2.3.3.4.4.....2014.2014}\)
\(-A=\frac{\left(1.2.3...2013\right).\left(3.4.5.6...2015\right)}{\left(2.3.4.5....2014\right).\left(2.3.4.5...2014\right)}=\frac{1.2015}{2.2014}=\frac{2015}{4028}\)
\(A=-\frac{2015}{4028}\)
Vậy.....
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-A=(\(1-\frac{1}{2^2}\)) . (\(1-\frac{1}{3^2}\))......(\(1-\frac{1}{2014^2}\))
-A= \(\frac{3}{4}\). \(\frac{8}{9}\). ...... \(\frac{4056195}{4056196}\)
-A= \(\frac{1.3.2.4.......2013.2015}{2.2.3.3.......2.14.2014}\)
-A= \(\frac{\left(1.2.3...2013\right)\left(3.4.5...2015\right)}{\left(2.3.4...2014\right)\left(2.3.4...2014\right)}\)
-A= \(\frac{2015}{2014.2}\)
-A=\(\frac{2015}{4028}\)
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