Tính
B\(=\frac{1}{1008\times2014}+\frac{1}{1009\times2013}+\frac{1}{1010\times2012}+...+\frac{1}{2014\times1008}\)
Ai đúng mình tick cho
Những câu hỏi liên quan
Tính tổng sau:\(A=\frac{1}{2016\times2015}+\frac{1}{2015\times2014}+\frac{1}{2014\times2013}+..........+\frac{1}{2}\)
Giúp mình làm nha, ai nhanh mình sẽ "tờ ích tích sắc tích" cho. Chúc các bạn học tốt!!!
Ta có :
\(A=\frac{1}{2016.2015}+\frac{1}{2015.2014}+\frac{1}{2014.2013}+...+\frac{1}{2}\)
\(A=\frac{1}{1.2}+...+\frac{1}{2013.2014}+\frac{1}{2014.2015}+\frac{1}{2015.2016}\)
\(A=\frac{1}{1}-\frac{1}{2}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2016}\)
\(A=1-\frac{1}{2016}\)
\(A=\frac{2015}{2016}\)
Vậy \(A=\frac{2015}{2016}\)
Chúc bạn học tốt ~
Đúng 0
Bình luận (0)
CMR :
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2013}-\frac{1}{2014}=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}\)
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}-1-\frac{1}{2}-...-\frac{1}{1007}\)
\(=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}\) (đpcm)
Đúng 0
Bình luận (0)
Cho S=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.........+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)
P=\(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...........+\frac{1}{2014}+\frac{1}{2015}\)
Tính (S-P)2016
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}+\frac{1}{2015}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1007}\right)\)
\(S=\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}\)
\(\Rightarrow\left(S-P\right)^{2016}=\left(\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}-\frac{1}{1008}-\frac{1}{1009}-...-\frac{1}{2015}\right)^{2016}=0^{2016}=0\)
Đúng 0
Bình luận (0)
Ta thấy:
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1007}\right)+\frac{1}{2015}\)
\(S=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}+\frac{1}{2015}\)
Mà \(P=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}+\frac{1}{2015}\) nên:
\(S=P\)\(\Rightarrow S-P=0\)\(\Rightarrow\left(S-P\right)^{2016}=0\)
Đúng 0
Bình luận (0)
Cho \(C=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+........-\frac{1}{2014}\)va \(D=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+........+\frac{1}{2014}\)tinh \(\left(\frac{C}{D}\right)^{2015}\)
cho hỏi: A=\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{2015\cdot2016}\)
B=\(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
Tính B-A
Ai giúp cảm ơn!
a) A = \(\frac{4024\times2014-2}{2011+2012\times2010}\)mình biết kết quả ý này bằng 2 bạn nào giải giùm rồi xem có đúng kq ko
b) B = \(\frac{2012\times2013+2014}{2010-2012\times2015}\)ý này bằng 1
A = \(\frac{4024x\left(2010+4\right)-2}{2011+2012x2010}\)= \(\frac{2024x2010+4024-2}{2011+2012x2010}\)=\(\frac{4024x2010+4022}{2011+2012x2010}\)= 2
câu B hình như sai đề bài . mk moi hoc lop 6 thoi nen cũng ko chắc .
Đúng 0
Bình luận (0)
Chứng minh rằng : \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2013.2014}=\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2014}\)
a)so sánh:sqrt{17}+sqrt{26}+1vàsqrt{99}b)chứng minh:frac{1}{sqrt{ }1}+frac{1}{sqrt{ }2}+frac{1}{sqrt{ }3}+...+frac{1}{sqrt{ }99}+frac{1}{sqrt{ }100}10c)cho:S1-frac{1}{2}+frac{1}{3}-frac{1}{4}+...+frac{1}{2013}-frac{1}{2014}+frac{1}{2015}vàPfrac{1}{1008}+frac{1}{1009}+frac{1}{1010}+...+frac{1}{2014}+frac{1}{2015}tính left(S-Pright)^{2016}
Đọc tiếp
a)so sánh:
\(\sqrt{17}+\sqrt{26}+1và\sqrt{99}\)
b)chứng minh:\(\frac{1}{\sqrt{ }1}+\frac{1}{\sqrt{ }2}+\frac{1}{\sqrt{ }3}+...+\frac{1}{\sqrt{ }99}+\frac{1}{\sqrt{ }100}>10\)
c)cho:S=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)vàP=\(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}+\frac{1}{2015}\)tính \(\left(S-P\right)^{2016}\)
Chứng minh :
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
Mau nha,chiều nộp rồi.AI làm mau nhất và đứng nhất ,mình tick cho