ta có: L = \(\frac{7}{3}+\frac{11}{3^2}+\frac{15}{3^3}+...+\frac{403}{3^{100}}\)
<=> \(3L=7+\frac{11}{3}+\frac{15}{3^2} +..+\frac{403}{3^{99}}\)
=> \(3L-L=\left(7+\frac{11}{3}+\frac{15}{3^2}+...+\frac{403}{3^{99}}\right)-\left(\frac{7}{3}+\frac{11}{3^2}+...+\frac{403}{3^{100}}\right)\)
<=> \(2L=7+\left(\frac{11}{3}-\frac{7}{3}\right)+\left(\frac{15}{3^2}-\frac{11}{3^2}\right)+...+\left(\frac{403}{3 ^{99}}-\frac{399}{3^{99}}\right)-\frac{403}{3^{100}}\)
<=> \(2L=7+4\cdot\frac{1}{3}+4\cdot\frac{1}{3^2}+..+4\cdot\frac{1}{3^{99}}-\frac{403}{3^{100}}\)
<=> \(2L=7+4\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\frac{403}{3^{100}}\)
<=>\(2L=7+4\left[\frac{1}{2}\cdot\left(1-\frac{1}{3^{99}}\right)\right]-\frac{403}{3^{100}}\)
<=> \(2L=7+2-\frac{2}{3^{99}}-\frac{403}{3^{100}}\)
<=> \(L=3,5+1-\frac{1}{3^{99}}-\frac{403}{2\cdot3^{100}}\)
<=> \(L=4,5-\frac{1}{3^{99}}-\frac{403}{2\cdot3^{100}}
