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Tham khảo:Câu hỏi của Hoang Phươngpsh - Toán lớp 5 - Học toán với OnlineMath
\(\frac{1}{1.2}\frac{1}{2.3}\frac{1}{3.4}...........\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
Ta đặt: A=\(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{100}}\)
\(\Rightarrow4A=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{99}}\)
4A - A = \(1-\frac{1}{4^{100}}\)= 3A = \(\frac{4^{100}-1}{4^{100}}\)\(\Rightarrow A=\frac{4^{100}-1}{4^{100}.3}=\frac{1}{3}-\frac{1}{4^{100}.3}=\frac{1}{3}-\frac{1}{4^{100}}.\frac{1}{3}=\frac{1}{3}\left(1-\frac{1}{4^{100}}\right)\)
hết
\(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right).\left(\frac{1}{4}-1\right)...\left(\frac{1}{100}-1\right)\)
\(=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-99}{100}\)
\(=\frac{1}{-2}.\frac{-2}{3}.\frac{3}{-4}...\frac{-99}{100}=\frac{1}{100}\)
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)......\left(\frac{1}{100}-1\right)\)
\(=\frac{-1}{2}\frac{-2}{3}\frac{-3}{4}......\frac{-99}{100}\)
\(=\frac{-1}{2}\frac{2}{-3}\frac{-3}{4}......\frac{99}{-100}\)
\(=\frac{-1}{-100}=\frac{1}{100}\)
a, a có 51 số ,chia thành 25 cặp mỗi cặp hai số hạng (trừ số 1)
Ta có (100-98)+(96-94)+...+(4-2)+1
=2.25+1
=51
b,1/2.2/3.3.4....2016/2017.2017/2018
1/2018
c,3/2.4/3....2018/2017
2018/2=1009
\(100-98+96-94+...+4-2+1\)
\(=2+2+...+2+1\)( có 50 ố 2 )
\(=2.50+1\)
\(=\)\(101\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2017}{2018}\)
\(=\frac{1}{2018}\)
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{2017}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{2018}{2017}\)
\(=\frac{2018}{2}\)
\(=1009\)
học tốt
=\(\frac{0}{2}\).\(\frac{0}{3}\).\(\frac{0}{4}\).........\(\frac{0}{100}\)
=0.0.0......0
=0
Ta có:
\(A=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
=> \(3A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
=> \(A+3A=1-\frac{1}{3^{100}}\)
=> \(4A=\frac{3^{100}-1}{3^{100}}\)
=> \(A=\frac{3^{100}-1}{4.3^{100}}\)
3/2 * 4/3 * 5/4 *...*101/100
101/2
t nha