ghep cac pso o sbc thanh tung cap de mc giong mau cua cac pso tuong ung ở so chia.bien doi sbc :cong tung pso cach deu 2 dau .
ta duoc\(\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+....+\left(\frac{1}{49}+\frac{1}{51}\right)=\\
\frac{100}{1.99}+\frac{100}{3.97}+....+\frac{100}{49.51}\)bieu thuc nay gap 50 lan so chia.suy ra =50! dung 100% do nha!
Cho A=\(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{49.51}}\) . Rút gọn A=
\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}=\frac{1}{k}\times\left(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\right)\)
Tìm giá trị của k.
1.chứng minh rằng A<\(\frac{1}{16}\) biết A=\(\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+.....+\frac{99}{5^{100}}\)
2.tính (M-N)\(^3\) biết:
M=1-\(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
N=\(\frac{1}{1010}+\frac{1}{1011}+.....+\frac{1}{2019}\)
\(\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+...+\frac{1}{99^2}< \frac{5}{4}\)
chứng minh rằng:
A=\(\frac{1}{5}+\frac{1}{14}+\frac{1}{28}+\frac{1}{44}+\frac{1}{61}+\frac{1}{85}+\frac{1}{97}< \frac{1}{2}\)
\(1:\frac{99}{100}:\frac{98}{99}:\frac{97}{98}:...:\frac{2}{3}:\frac{1}{2}\)
tính
Bài 1:So sánh Avà B biết rằng:
A=\(\frac{10^{15}+1}{10^{16}+1};\) B=\(\frac{10^{16}+1}{10^{17}+1}\)
A=\(\frac{3}{8^3}+\frac{7}{8^4}\); B=\(\frac{7}{8^3}+\frac{3}{8^4}\)
A=\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+.......+\frac{1}{19}+\frac{1}{20};\) B=\(\frac{1}{2}\)
Bài 2:Dạng tính tổng đặc biệt:
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+.....+\frac{1}{99\cdot100}\)
\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+.....+\frac{2}{99\cdot101}\)
\(C=\frac{3^2}{10}+\frac{3^2}{40}+\frac{3^2}{88}+......+\frac{3^2}{340}\)
\(D=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+......+\frac{1}{3^8}\)
\(E=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{99}\right)\)
Bài 3:Dạng chứng minh:
\(A=1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{99}.\)Chứng minh rằng A chia hết cho 100
A=\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{70}\).Chứng minh rằng A>\(\frac{4}{3}\)
Chứng Minh Rằng
a) \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b) \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+.....+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
a=1+\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}
