Chung to rang
b) B = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}< \frac{1}{2}\)
2, chung minh rang
a, \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}<\frac{1}{3}\)
b,\(\frac{1}{3}-\frac{2}{^{3^2}}+\frac{3}{3^4}+........+\frac{99}{3^{99}}-\frac{100}{3^{100}}<\frac{3}{16}\)
mình chỉ gợi ý thôi, vì viết cái này mỏi tay lắm thông cảm nha
Ở phần ''a'' bạn hãy đổi ra thành:2=2;4=2;.....sau dó bạn CM \(\frac{1}{2^2}<\frac{1}{1.2}.....\) rồi hãy suy ra nhỏ hơn \(\frac{1}{3}\)
còn phần ''b'' bạn hãy tách ra nha
à chỗ 2=2;4=2 bạn sửa thành : \(2=2^1;4=2^2\) nhé
Chứng minh rằng:
a. \(\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\frac{4}{3^5}+...+\frac{99}{3^{100}}+\frac{100}{3^{101}}< \frac{1}{4}\)
b.\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
c.\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{1}{16}\)
d. \(\frac{1}{5^2}-\frac{2}{5^3}+\frac{3}{5^4}-\frac{4}{5^5}+...+\frac{99}{5^{100}}-\frac{100}{5^{101}}< \frac{1}{36}\)
a)A= \frac{1}{3^1}311+\frac{1}{3^2}321+\frac{1}{3^3}331+.........+\frac{1}{3^{99}}3991
b)B=\frac{1}{3^1}311+\frac{2}{3^2}322+\frac{3}{3^3}333+..........+\frac{99}{3^{99}}39999
các bạn làm hộ mình nhé
Chung minh rang: \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+....+\frac{99}{3^{99}}-\frac{100}{3^{100}}<\frac{3}{16}\)
Dat A=1/3-2/32+3/33-4/34+...+99/399-100/3100
3A=1-2/3+3/32-4/33+...+99/398-100/399
3A+A=1-1/3+1/32-1/33+...+1/398-1/399-100/3100=4A
4A.3=3-1+1/3-1/32+...+1/397-1/398-100/399=12A
4A+12A=3-100/399-1/399-100/3100
16A=3-300/3100-3/3100-100/3100=3-403/3100<3
A<3/16
Chung to...
cho C=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)chung minh rang C>\(\frac{1}{2}\)
CMR:
a, \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b, \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+.....+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
CMR
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) \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+.....+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
Chứng minh :
a) \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\) \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{4^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
b)\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{79}+\frac{1}{80}< \frac{7}{12}\)
c) Cho \(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)
Chứng minh \(1< S< 2\)