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A = \(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)
3A= \(1+\frac{1}{3}+...+\frac{1}{3^{2006}}+\frac{1}{3^{2007}}\)
3A-A= \(1-\frac{1}{3^{2008}}\)
a) \(A=1+2+2^2+...+2^{2016}\)
\(\Rightarrow2A=2+2^2+2^3+...+2^{2017}\)
\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{2017}\right)-\left(1+2+2^2+...+2^{2016}\right)\)
\(\Rightarrow A=2^{2017}-1\)
Vậy \(A=2^{2017}-1\)
b) \(B=1.2.3+2.3.4+...+n\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow4B=1.2.3.4+2.3.4\left(5-1\right)+...+n\left(n+1\right)\left(n+2\right)\left[\left(n+3\right)-\left(n-1\right)\right]\)
\(\Rightarrow4B=1.2.3.4+2.3.4.5-1.2.3.4+...+n\left(n+1\right)\left(n+2\right)\left(n+3\right)-\left(n-1\right)n\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow4B=n\left(n+1\right)\left(n+2\right)\left(n+3\right)\)
\(\Rightarrow B=\frac{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}{4}\)
Vậy...
4A = 4.[1.2.3 + 2.3.4 + 3.4.5 + … + (n – 1).n.(n + 1)]
4A = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + … + (n – 1).n.(n + 1).4
4A = 1.2.3.4 + 2.3.4.(5 – 1) + 3.4.5.(6 – 2) + … + (n – 1).n.(n + 1).[(n + 2) – (n – 2)]
4A = 1.2.3.4 + 2.3.4.5 – 1.2.3.4 + 3.4.5.6 – 2.3.4.5 + … + (n – 1).n(n + 1).(n + 2) – (n – 2).(n – 1).n.(n + 1)
4A = (n – 1).n(n + 1).(n + 2)
A = (n – 1).n(n + 1).(n + 2) : 4.
4B = 1.2.3.4 + 2.3.4.4 + ... + (n - 1)n(n + 1).4
= 1.2.3.4 - 0.1.2.3 + 2.3.4.5 - 1.2.3.4 + ... + (n - 1)n(n + 1)(n + 2) - [(n - 2)(n - 1)n(n + 1)]
= (n - 1)n(n + 1)(n + 2) - 0.1.2.3 = (n - 1)n(n + 1)(n + 2)
\(S_n=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(2S_n=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}\)
\(2S_n=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(2S_n=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(S_n=\frac{1}{4}-\frac{1}{2\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)=\frac{n\left(n+3\right)}{4\left(n+1\right)\left(n+2\right)}\)
