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Chứng minh rằng :
\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+..+\frac{1}{3^{2014}}\) \(< \frac{1}{2}\)
Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2014}}\)=>\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2013}}\)
=>\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2013}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2014}}\right)\)
=>\(2A=1-\frac{1}{2^{2014}}< 1\Rightarrow A< \frac{1}{2}\)(đpcm)
B=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+.....+\frac{1}{3^{2012}}+\frac{1}{3^{2013}}\)
3B=\(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^{2011}}+\frac{1}{3^{2012}}\)
3B-B=\(\left(1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{2011}}+\frac{1}{3^{2012}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{2012}}+\frac{1}{3^{2013}}\right)\)
2B=\(1-\frac{1}{3^{2013}}\)
\(\Rightarrow2B< 1\)
\(\Rightarrow B< \frac{1}{2}\)
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2013}}\)
\(3B=\frac{1}{3}.3+\frac{1}{3^2}.3+\frac{1}{3^3}.3+...+\frac{1}{3^{2013}}.3\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2012}}\)
\(3B-B=2B=\)
3B= \(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2012}}\)
B= \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2012}}+\frac{1}{3^{2013}}\)
2B= 1 + 0 + 0 + 0 +.......+ 0 - \(\frac{1}{3^{2013}}\)
\(\Rightarrow2B=1-\frac{1}{3^{2013}}\)
\(\Rightarrow B=\frac{1}{2}-\frac{1}{2.3^{2013}}\)
\(\Rightarrow B< \frac{1}{2}\)
Vậy \(B< \frac{1}{2}\).
Ta có :
M = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
3M = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
3M - M = ( \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)) - ( \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\))
2M = \(1-\frac{1}{3^{99}}< 1\)
\(\Rightarrow M=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)
3M=1+1/3+1/3^2+....+1/3^98
2M=3M-M=(1+1/3+1/3^2+....+1/3^98)-(1/3+1/3^2+....+1/3^99) = 1-1/3^99 < 1
=> M < 1/2
=> ĐPCM
k mk nha
sửa đề câu 1 :
\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)
\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{100-1}{100!}\)
\(=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=1-\frac{1}{100!}< 1\)
sửa đề câu 2
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)
\(3C=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{98}}\)
\(2C=3C-C=1-\frac{1}{3^{99}}\Rightarrow C=\left(1-\frac{1}{3^{99}}\right):2=\frac{1}{2}-\frac{1}{2.3^{99}}< \frac{1}{2}\)
Đặt M=1/3+1/3^2+....+1/3^99
Ta có:1/(3^n)+1/(3^(n+1))=2/(3^(n+1))(cái này bạn tự quy đồng ra ra nhé!).
Áp dụng ta có:1-1/3=2/3
1/3-1/(3^2)=2/(3^2)
1/(3^2)-1/(3^3)=2/(3^3)
....
1/(3^98)-1/(3^99)=2/(3^99).
Cộng từng vế các phép tính với nhau ta có:1-1/(3^99)=2M.
Mà 1-1/(3^99)<1 nên 2M<1 nên M<1/2(điều phải chứng minh)
\(M=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2013}}+\frac{1}{5^{2014}}\)
\(5M=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2013}}\)
\(\Rightarrow4M=1-\frac{1}{5^{2014}}< 1\)
\(\Rightarrow M< \frac{1}{4}< \frac{1}{3}\)
3C = 1+1/3+1/3^2+....+1/3^98
2C = 3C - C = (1+1/3+1/3^2+...+1/3^98) - (1/3+1/3^2+1/3^3+...+1/3^99) = 1- 1/3^99 < 1
=> C < 1/2
k mk nha
Đặt \(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2014}}\)
\(\Rightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2013}}\\ \Rightarrow2A=1-\dfrac{1}{3^{2014}}\\ \Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{2014}}< \dfrac{1}{2}\)