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ta có:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+..+\frac{19}{9^2.10^2}=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{10^2-9^2}{9^2.10^2}\)
\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}=1-\frac{1}{10^2}
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn chép đề sai rồi, mình sửa lại đề và làm luôn nhé :
Ta có :
\(D=\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.305.308}\)
\(\Rightarrow D=\frac{5}{6}.\left(\frac{6}{5.8.11}+\frac{6}{8.11.14}+...+\frac{6}{302.305.308}\right)\)
\(\Rightarrow D=\frac{5}{6}.\left(\frac{1}{5.8}-\frac{1}{8.11}+\frac{1}{8.11}\frac{1}{11.14}+...+\frac{1}{302.305}-\frac{1}{305.308}\right)\)
\(\Rightarrow D=\frac{5}{6}.\left(\frac{1}{5.8}-\frac{1}{305.308}\right)\)
\(\Rightarrow D=\frac{5}{6}.\frac{1}{40}-\frac{5}{6}.\frac{1}{305.308}\)
\(\Rightarrow D=\frac{1}{48}-\frac{5}{6.305.308}< \frac{1}{48}\) (đpcm )
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\(B=1+5+5^2+5^3+...+5^{2008}+5^{2009}\)
\(\Rightarrow 5B=5+5^2+5^3+5^4+...+5^{2009}+5^{2010}\)
Trừ theo vế:
\(5B-B=(5+5^2+5^3+5^4+...+5^{2009}+5^{2010})-(1+5+5^2+...+5^{2009})\)
\(4B=5^{2010}-1\)
\(B=\frac{5^{2010}-1}{4}\)
\(S=\frac{3^0+1}{2}+\frac{3^1+1}{2}+\frac{3^2+1}{2}+..+\frac{3^{n-1}+1}{2}\)
\(=\frac{3^0+3^1+3^2+...+3^{n-1}}{2}+\frac{\underbrace{1+1+...+1}_{n}}{2}\)
\(=\frac{3^0+3^1+3^2+..+3^{n-1}}{2}+\frac{n}{2}\)
Đặt \(X=3^0+3^1+3^2+..+3^{n-1}\)
\(\Rightarrow 3X=3^1+3^2+3^3+...+3^{n}\)
Trừ theo vế:
\(3X-X=3^n-3^0=3^n-1\)
\(\Rightarrow X=\frac{3^n-1}{2}\). Do đó \(S=\frac{3^n-1}{4}+\frac{n}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
D<1/2.3+1/3.4+...+1/(n-1).n
<=>D<1/2-1/3+1/3-1/4+...+1/(n-1)-1/n
<=>D<1/2-1/n
<=>D<1(n>0)