Ta có:25C=52C=\(1+\dfrac{1}{5^2}+\dfrac{1}{5^4}+...+\dfrac{1}{5^{98}}\)
=>25C-C=(\(1+\dfrac{1}{5^2}+\dfrac{1}{5^4}+...+\dfrac{1}{5^{98}}\))-(\(\dfrac{1}{5^2}+\dfrac{1}{5^4}+...+\dfrac{1}{5^{100}}\))
=>24C=1-\(\dfrac{1}{5^{100}}=\dfrac{5^{100}-1}{5^{100}}\)
=>C=\(\dfrac{5^{100}-1}{24.5^{100}}\)
Vậy C=\(\dfrac{5^{100}-1}{24.5^{100}}\)
Thu gọn các tổng sau:
a. A=8.5100.(\(\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{100}}\)) +1
b. B=\(\dfrac{4}{3}-\dfrac{4}{3^2}+...-\dfrac{4}{3^{100}}\)
Thu gọn :
C = \(\dfrac{\dfrac{1}{2}-\dfrac{1}{2}:\dfrac{3}{4}-\dfrac{3}{4}}{\dfrac{2}{3}-\dfrac{2}{3}:\dfrac{5}{6}-\dfrac{5}{6}}\)
Thu gọn:D= \(\dfrac{1}{5}\)-\(\dfrac{1}{5^2}\)+\(\dfrac{1}{5^3}\)-\(\dfrac{1}{5^4}\)+\(\dfrac{1}{5^5}\)-...- \(\dfrac{1}{5^{100}}\)+\(\dfrac{1}{5^{101}}\)
Chứng minh rằng:
1) B =\(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{19}>1\)
2) \(A=\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+\dfrac{4}{5^4}+...+\dfrac{500}{5^{500}}<100\)
3) \(C=\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+\dfrac{1}{5^3}+...+\dfrac{1}{500^3}<\dfrac{1}{4}\)
4) \(D=\dfrac{4}{3}+\dfrac{10}{9}+\dfrac{28}{27}+...+\dfrac{3^{98}+1}{3^{98}}<100\)
Làm giúp mình sớm nha! Thanks.
Thu gọn:
A = \(\dfrac{2^4.3^3+2^3.3^4}{2^5.3^3-2^4.3^2}\) D = \(\dfrac{\left(\dfrac{1}{2}-\dfrac{1}{3}\right).\left(\dfrac{2}{3}-\dfrac{3}{4}\right)}{\left(\dfrac{1}{2}+\dfrac{1}{3}\right).\left(\dfrac{2}{3}+\dfrac{3}{4}\right)}\)
B = \(\dfrac{2^3-3^4-2^4.3^3}{2^5.3^4-2^6.3^3}\) E = \(\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{3}{4}\right).\left(\dfrac{2}{3}-\dfrac{3}{4}+\dfrac{5}{6}\right)\)
C = \(\dfrac{\dfrac{1}{2}-\dfrac{1}{2}:\dfrac{3}{4}-\dfrac{3}{4}}{\dfrac{2}{3}-\dfrac{2}{3}:\dfrac{5}{6}-\dfrac{5}{6}}\)
Giúp mình với ! 10k nha
Tính nhanh
A=\(\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{1}{8}\right)+...+\left(1-\dfrac{1}{1024}\right)\)
B=4.5100 .\(\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{100}}\right)+1\)
CMR C = \(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{9999}{10000}< \dfrac{1}{100}\)
bài 4 : cmr :
c) C = \(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}....\dfrac{9999}{10000}< \dfrac{1}{100}\)
K=\(\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}\)
CMR:\(\dfrac{1}{5}< K< \dfrac{1}{3}\)
