Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
H = 2012 - 1 - ( \(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+99}\))
= 2011 - ( \(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{\left(99+1\right).\left[\left(99-1\right):1+1\right]:2}\)
= 2011 - ( \(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{4950}\))
= 2011 - 2.( \(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\))
= 2011 - 2.(\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\))
= 2011 - 2.( \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\))
= 2011 - 2.(\(\frac{1}{2}-\frac{1}{100}\)) = 2011 - 2.\(\frac{49}{100}\)= 2011 - \(\frac{49}{50}\)= \(\frac{100501}{50}\)
\(H=2012-\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+99}\right)\)
\(=2012-\left(1+\frac{1}{2\left(2+1\right):2}+\frac{1}{3\left(3+1\right):2}+...+\frac{1}{99\left(99+1\right):2}\right)\)
\(=2012-\left(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\right)\)
\(=2012-2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{2}{99.100}\right)\)
\(=2012-2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2012-2\left(1-\frac{1}{100}\right)\)
\(=2012-2\cdot\frac{99}{100}\)
\(=2012-\frac{99}{50}\)
\(=\frac{100501}{50}\)
\(\frac{-1}{9}.\frac{2}{5}-\frac{2}{9}.\frac{-1}{6}-\frac{29}{9}.\frac{1}{15}\)=\(\frac{-2}{45}-\frac{-1}{17}-\frac{29}{135}\)
* Tính BCNN (45, 17, 135)
45= 32. 5
17= 17
135= 33. 5
BCNN (45, 17, 135)= 33. 5. 17= 2295
* Tìm thừa số phụ
2295: 45= 51
2295: 17= 135
2295: 135= 17
* Ta có:
\(\frac{-2}{45}-\frac{-1}{17}-\frac{29}{135}\)= \(\frac{-102-\left(-135\right)-493}{2295}\)= \(\frac{-460}{2295}\)
Mình làm thế thôi, kết quả cuối bạn tự rút gọn nha!
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+......+\frac{1}{2^{2014}}\)
\(\Rightarrow A<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{2.3}+.....+\frac{1}{2013.2014}\)
\(\Rightarrow A<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2013}-\frac{1}{2014}\)
\(\Rightarrow A<\frac{1}{1}-\frac{1}{2014}\)
\(\Rightarrow A<1\)
**** cho mình nha ! ^_^
A < 1
xin lỗi mình không biết cách viết phân số!!!!
nha!!!!
\(\frac{1}{2!}+\frac{2!}{4!}+...+\frac{198!}{200!}=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-...+\frac{1}{199}-\frac{1}{200}=\left(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
(50-1):1+1=50 số
=(50-49)+(48-47)+...+(4-3)+(2-1). Ta có 25 cặp số
=1+1+1+....+1
=1.25
=25
Đặt M=\(\frac{A}{B}\)
A=1+2+22+23+.....+22012
2A=2+22+23+......+22013
2A-A=(2+22+23+....+22013) - (1+2+22+.....+22012)
A=22013 - 1
B=22014-2
B=2.(22013-1)
=>M=\(\frac{2^{2013}-1}{2.\left(2^{2013}-1\right)}\)=\(\frac{1}{2}\)