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C = 1/3 + 1/3^2 + 1/3^3 + ... =1/3^99
=> C = 1/3^99 = 1/(3^99)
=> C < 1/2 (đpcm)
2A=2^101-2^100+2^98+...+2^3-2^2
3A = 2A + A
3A = 2^101 - 2 ( Cứ tính là ra , âm vs dương triệt tiêu )
A = (2^101-2) :3
B tăng tự
\(\frac{T}{M}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{1}{99}+\frac{2}{98}+...+\frac{98}{2}+\frac{99}{1}}\)
Xét M - 99 + 98 = \(\frac{100}{99}+\frac{100}{98}+...+\frac{100}{2}\)
\(\Leftrightarrow M-1=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)\)
\(\Rightarrow M=\frac{100}{100}+100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(\Rightarrow\frac{T}{M}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)}=\frac{1}{100}\)
\(M=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\cdot\cdot\cdot\frac{9999}{10000}\)
\(=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot\cdot\cdot\cdot\frac{99.101}{100.100}\)
\(=\frac{1}{2}\cdot\frac{101}{100}=\frac{101}{200}\)
Xét vế phải :
\(VP=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}-1\)
\(=2.\left(\frac{99}{100}-\frac{97}{98}+...+\frac{7}{8}-\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)
\(=2\left[\left(1-\frac{1}{100}\right)-\left(1-\frac{1}{98}\right)+...+\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{2}\right)\right]\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}+\frac{1}{26}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}=VT\Rightarrow\left(đpcm\right)\)
A = ( 4/4 + 2/3 ) - ( 51/3 - 6/5 ) - ( 6 - 7/4 + 3/2 )
Sau đó quy đồng rồi trừ cả là đc
B tương tự
C=13/15
D cx thế . Bạn tự vận dụng đi . Xl vì ko giải đc . Mik đang gấp
a
\(A=1+3+3^2+3^3+....+3^{100}\)
\(3A=3+3^2+3^3+3^4+.....+3^{101}\)
\(2A=3^{101}-1\)
\(A=\frac{3^{101}-1}{2}\)
b
\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(B=1-\frac{1}{2^{99}}\)
c
\(C=5^{100}-5^{99}+5^{98}-5^{97}+....+5^2-5+1\)
\(5C=5^{101}-5^{100}+5^{99}-5^{98}+....+5^3-5^2+5\)
\(6C=5^{101}+1\)
\(C=\frac{5^{101}+1}{6}\)
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\)
\(\Rightarrow\frac{1}{2}B=\)\(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{100}\)
\(\Rightarrow B-\frac{1}{2}B=\left[\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\right]-\left[\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{100}\right]\)
\(\Rightarrow\frac{1}{2}B=\frac{1}{2}-\left(\frac{1}{2}\right)^{100}\Rightarrow B=\left[\frac{1}{2}-\left(\frac{1}{2}\right)^{100}\right].2\)