cmr:a, 1/2+2/3+3/4+...+99/100<1
b, cmr mọi số nguyên dương n thì:3^n+2 - 2^n+2 + 3^n - 2^n chia hết cho 10
AI NHANH NHẤT MK TICK CHO! THANKS >-<
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LH
0
2 tháng 4 2020
Câu hỏi của Phùng Tuệ Minh - Toán lớp 7 - Học toán với OnlineMath
SY
19 tháng 3 2016
a)Đặt A= \(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{8}\) - \(\frac{1}{16}\) + \(\frac{1}{32}\) - \(\frac{1}{64}\) => A=\(\frac{1}{2^1}\) - \(\frac{1}{2^2}\) + \(\frac{1}{2^3}\) - \(\frac{1}{2^4}\) + \(\frac{1}{2^5}\) - \(\frac{1}{2^6}\)
=> 2A= 1-\(\frac{1}{2^1}\) + \(\frac{1}{2^2}\) - \(\frac{1}{2^3}\) + \(\frac{1}{2^4}\) - \(\frac{1}{2^5}\)
=> 3A= 1- \(\frac{1}{2^6}\) <1 => A<\(\frac{1}{3}\) => đpcm.
b) Đặt B=\(\frac{1}{3}\) - \(\frac{2}{3^2}\) + \(\frac{3}{3^3}\) - \(\frac{4}{3^4}\) +..+ \(\frac{99}{3^{99}}\) - \(\frac{100}{3^{100}}\)
=> 3B=1-\(\frac{2}{3}\) + \(\frac{3}{3^2}\) - \(\frac{4}{3^3}\) +...+\(\frac{99}{3^{98}}\) - \(\frac{100}{3^{99}}\)
=> 4B= 1-\(\frac{1}{3}\) + \(\frac{1}{3^2}\) - \(\frac{1}{3^3}\) +...+\(\frac{1}{3^{99}}\) - \(\frac{100}{3^{99}}\) < 1-\(\frac{1}{3}\) + \(\frac{1}{3^2}\) - \(\frac{1}{3^3}\) +...+\(\frac{1}{3^{99}}\) (1)
Đặt B= 1-\(\frac{1}{3}\) + \(\frac{1}{3^2}\) - \(\frac{1}{3^3}\) +...+\(\frac{1}{3^{99}}\)
=> 3B= 3-1+\(\frac{1}{3}\) - \(\frac{1}{3^2}\) + \(\frac{1}{3^3}\) - \(\frac{1}{3^4}\) +...+ \(\frac{1}{3^{98}}\)
=> 4B= 3-\(\frac{1}{3^{99}}\) <3 => B<\(\frac{3}{4}\) (2)
=> 4A<B<\(\frac{3}{4}\) => A<\(\frac{3}{16}\) => đpcm.
IL
0
NT
7 tháng 3 2018
T làm biếng lắm; làm C thôi
\(A=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\\ \Rightarrow A< \dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{100}{101}\\ \Rightarrow A^2< \left(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\right).\left(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{100}{101}\right)\\ =\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{99}{100}.\dfrac{100}{101}\\ =\dfrac{1}{101}< \dfrac{1}{100}\\ \Rightarrow A< \dfrac{1}{10}\)
Làm tương tự ta được A > 1/15
9 tháng 3 2018
câu a
\(A=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{30}>\dfrac{20}{30}=\dfrac{2}{3}>\dfrac{1}{3}\)
\(A=\left(\dfrac{1}{11}+..+\dfrac{1}{15}\right)+\left(\dfrac{1}{16}+...+\dfrac{1}{30}\right)< 5.\dfrac{1}{10}+25.\dfrac{1}{15}=\dfrac{1}{2}+\dfrac{5}{3}=\dfrac{8}{6}=\dfrac{4}{3}< \dfrac{5}{2}\)
PT
3
NM
14 tháng 2 2016
A=1+4+42+43+...+499
=>4A=4+42+43+44+...+4100
=>4A-A=(4+42+43+44+...+4100)-(1+4+42+43+...+499)
=>3A=4100-1
=>A=\(\frac{4^{100}-1}{3}\) < 4100
=>A<B
T1
14 tháng 2 2016
\(A=1+4+4^2+4^3+...+4^{99}\)
=> \(4A=4+4^2+4^3+4^4+...+4^{100}\)
=> \(4A-A=\left(4+4^2+4^3+...+4^{100}\right)-\left(1+4+4^2+...+4^{99}\right)\)
=> \(3A=4^{100}-1\)
=> \(A=\frac{4^{100}-1}{3}\)
Ta có : \(B=4^{100}\) => \(\frac{B}{3}=\frac{4^{100}}{3}\)
Vì \(4^{100}-1<4^{100}\) => \(\frac{4^{100}-1}{3}<\frac{4^{100}}{3}\) => \(A<\frac{B}{3}\) (đpcm)
XO
15 tháng 2 2020
Ta có : A = \(\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{99}{5^{100}}\)
=> 5A = \(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{99}{5^{99}}\)
=> 5A - A = \(\left(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{99}{5^{99}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{99}{5^{100}}\right)\)
=> 4A \(=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)
=> 20A = \(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}-\frac{99}{5^{99}}\)
Lấy 20A trừ A ta có :
20A - A = \(\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}-\frac{99}{5^{99}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}-\frac{99}{5^{100}}\right)\)
16A = \(1-\frac{99}{5^{99}}+\frac{99}{5^{100}}=1+99\left(\frac{1}{5^{100}}-\frac{1}{5^{99}}\right)=1-\frac{99.4}{5^{100}}\)
=> A = \(\frac{1}{16}-\frac{99}{4.5^{100}}< \frac{1}{16}\left(\text{ĐPCM}\right)\)
15 tháng 2 2020
Ta có :A=\(\frac{1}{5^2}+\frac{2}{5^3}+.....+\frac{99}{5^{100}}\)
5A=\(\frac{1}{5}+\frac{2}{5^2}+.....+\frac{99}{5^{99}}\)
5A -A=\(\left(\frac{1}{5}+\frac{2}{5^2}+...+\frac{99}{5^{99}}\right)\)-\(\left(\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{99}{5^{100}}\right)\)
4A =\(\frac{1}{5}+\frac{1}{5^2}+....+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)
Đặt B=\(\frac{1}{5}+\frac{1}{5^2}+.....+\frac{1}{5^{99}}\)
5B=\(1+\frac{1}{5}+...+\frac{1}{5^{98}}\)
5B - B =\(\left(1+\frac{1}{5}+...+\frac{1}{5^{98}}\right)\)- \(\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\right)\)
4B =\(1-\frac{1}{5^{99}}\)
Ta có :4A = B -\(\frac{99}{5^{100}}\)
16A = 4B -\(\frac{4.99}{5^{100}}\)=\(1-\frac{1}{5^{99}}-\frac{4.99}{5^{100}}\)
A = \(\frac{1}{16}-\frac{1}{5^{99}.16}-\frac{99}{5^{100}.4}\)< \(\frac{1}{16}\)
Suy ra: A <\(\frac{1}{16}\)
ai ủng hộ 9 li-ke tròn 100 Điểm hỏi đáp , thanks trước nha