Question

1) tính \(S=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{19683}\)

2) tính \(S=1+\dfrac{1}{5}+\dfrac{1}{25}+...+\dfrac{1}{78125}\)

Answer 1

1, Ta có \(\dfrac{\dfrac{1}{3}}{1}=\dfrac{1}{3};\dfrac{\dfrac{1}{9}}{\dfrac{1}{3}}=\dfrac{1}{3};...\)

-> Là cấp số nhân, q = 1/3 

Ta có \(S_9=1.\dfrac{1-\left(\dfrac{1}{3}\right)^9}{1-\left(\dfrac{1}{3}\right)}\approx1,5\)

b, Ta có \(\dfrac{\dfrac{1}{5}}{1}=\dfrac{1}{5};\dfrac{\dfrac{1}{25}}{\dfrac{1}{5}}=\dfrac{1}{5};...\)

-> Là cấp số nhân, q = 1/5 

\(S_7=\dfrac{1-\left(\dfrac{1}{5}\right)^7}{1-\dfrac{1}{5}}\approx1,25\)