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a)
\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-...-\frac{1}{64}=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-...-\frac{1}{2^6}=A\)
2A = 1 - \(\frac{1}{2}+\frac{1}{2^2}-...-\frac{1}{2^5}\)
2A + A = 1 - \(\frac{1}{2}+\frac{1}{2^2}-...-\frac{1}{2^5}+\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}-...-\frac{1}{2^6}\)
3A = \(1-\frac{1}{2^6}=\frac{2^6-1}{2^6}\)(đpcm)
\(\frac{31}{2}\)\(.\)\(\frac{32}{2}\)\(.\)\(\frac{33}{2}\)\(....\)\(\frac{60}{2}\)
\(=\)\(\left[\left(31.32.33....60\right)\right]\)\(.\)\(\left(\frac{1.2.3....30}{2^{30}}\right)\)\(.\)\(\left(1.2.3....30\right)\)
\(=\)\(\left[\frac{\left(1.3.5....59\right).\left(2.4.6....60\right)}{2.4.6....60}\right]\)\(=\)\(1.3.5....59\)
Vậy \(\frac{31}{2}\)\(.\)\(\frac{32}{2}\)\(.\)\(\frac{33}{2}\)\(....\)\(\frac{60}{2}\)\(=\)\(1.3.5....59\)
ta có:Đặt A= \(1.3.5.....59=\frac{1.2.3.4.....59.60}{2.4.6.....60}\)
=\(\frac{1.2.3.....59.60}{2^{30}.\left(1.2.3.....30\right)}=\frac{31.32.....59.60}{2^{30}}\)
= \(\frac{31}{2}.\frac{32}{2}.....\frac{59}{2}.\frac{60}{2}\)
vì \(\frac{31}{2}.\frac{32}{2}.....\frac{59}{2}.\frac{60}{2}\) = \(\frac{31}{2}.\frac{32}{2}.....\frac{59}{2}.\frac{60}{2}\)
\(\Rightarrow\)A= \(\frac{31}{2}.\frac{32}{2}.....\frac{59}{2}.\frac{60}{2}\)
( Điều phải chứng minh)
toán nâng cao lớp 6 đấy bạn nha
\(3B=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(B=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow4B=3B+B=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
+ Đặt \(M=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(3M=3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
\(\Rightarrow4M=3M+M=3-\frac{1}{3^{99}}\)
\(\Rightarrow M=\frac{3}{4}-\frac{1}{3^{99}\cdot4}\)
\(\Rightarrow4B=M-\frac{100}{3^{100}}=\frac{3}{4}-\frac{1}{3^{99}\cdot4}-\frac{100}{3^{100}}\)
\(\Rightarrow B=\frac{3}{16}-\frac{1}{3^{99}\cdot16}-\frac{100}{3^{100}\cdot4}\) \(\Rightarrow B< \frac{3}{16}\)
a) \(2A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)
\(A=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
\(\Rightarrow3A=2A+A=1-\frac{1}{2^6}\)
\(\Rightarrow A=\frac{1}{3}-\frac{1}{2^6\cdot3}< \frac{1}{3}\) ( đpcm )
Câu hỏi của Quỳnh Anh - Toán lớp 6 - Học toán với OnlineMath
Em tham khảo câu 1 2 cách 2 bạn hướng dẫn nhé!
a, \(1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}.......1\frac{1}{99}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}......\frac{10^2}{9.11}\)
\(=\frac{\left(2.3.4......10\right)\left(2.3.4....10\right)}{\left(1.2.3....9\right)\left(3.4.5....11\right)}\)
\(=\frac{10.2}{1.11}=\frac{20}{11}\)
b, Gọi A = \(\frac{31}{2}\cdot\frac{32}{2}\cdot\frac{33}{2}\cdot\cdot\cdot\cdot\frac{60}{2}\),gọi B = \(1.3.5....59\)
Ta có: \(A=\frac{31}{2}.\frac{32}{2}.\frac{33}{2}.....\frac{60}{2}\)
\(=\frac{31.32.33....60}{2^{30}}\)
\(=\frac{\left(31.32.33.....60\right)\left(1.2.3....30\right)}{2^{30}.\left(1.2.3....30\right)}\)
\(=\frac{1.2.3.....60}{\left(2.1\right)\left(2.2\right)\left(2.3\right)....\left(2.30\right)}\)
\(=\frac{1.2.3.....60}{2.4.6....60}\)
\(=\frac{\left(1.3.5...59\right)\left(2.4....60\right)}{2.4.6...60}\)
\(=1.3.5....59=B\)
Vậy A = B