\(=\frac{101}{101}\times\frac{100}{99}\times\frac{99}{98}\times\frac{98}{97}\times......\times\frac{2}{1}=\frac{101\times100\times99\times....\times2}{101\times99\times98\times97\times.....1}=100\)
C=\(\frac{101+100+99+98+...+3+2+1}{101-100+99-98+...+3-2+1}\)
D=\(\frac{3737.43-4343.37}{2+4+6+...+100}\)
Cho M =\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\) .Hãy chứng minh M<\(\frac{3}{16}\)
Câu 2 Chứng minh rằng :
\(\frac{1}{7^2}-\frac{1}{7^4}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}< \frac{1}{50}\)
Tính nhanh :
a) 17 x 8 + 51 x 4
b) 2 x 2 x 3 x 5 x 19
c) 54 x 275 + 825 x 15 + 275
d) 100 - 99 + 98 - 97 + 96 - 95 + 94 - 93 + ... + 4 - 3 + 2
e) \(\frac{167x198+98}{198x168-100}\)
g) \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{2019x2020}\)
h)\(\frac{4}{2x4}+\frac{4}{4x6}+\frac{4}{6x8}+...+\frac{4}{16x18}\)
k) 1,5 + 2,5 + 3,5 + 4,5 + 5,5 + 6,5 + 7,5 + 8,5
m) \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)
n) \(\frac{13}{50}+9\%+\frac{14}{100}+24\%\)
\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}=\frac{1}{k}\times\left(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\right)\)
Tìm giá trị của k.
So sánh A và B:
a) A = \(\frac{10^{19}+1}{10^{20}+1}\); B = \(\frac{10^{20}+1}{10^{21}+1}\)
b) A = \(\frac{9^{99}+1}{9^{100}+1}\); B = \(\frac{10^{98}-1}{10^{99}-1}\)
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) \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+.....+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
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)\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
Nhanh lên nhé! Mk đang cần gấp.
a=1+\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}
Bài 1: Chứng minh rằng: \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
Bài 2: Cho \(n\in N;n>1\). Chứng minh rằng: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{\left(n-1\right)^2}+\frac{1}{n^2}\notin N\)
