Sn=a(rn−1)r−1=1⋅(2n−1)2−1=2n−1S_n = \frac{a(r^n - 1)}{r - 1} = \frac{1 \cdot (2^n - 1)}{2 - 1} = 2^n - 1Sn=r−1a(rn−1)=2−11⋅(2n−1)=2n−1 ∑n=1∞Sn4n=∑n=1∞2n−14n=∑n=1∞(12)n−∑n=1∞(14)n=121−12−141−14=12−1−14−1=1−13=23\sum_{n=1}^{\infty} \frac{S_n}{4^n} = \sum_{n=1}^{\infty} \frac{2^n - 1}{4^n} = \sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n - \sum_{n=1}^{\infty} \left(\frac{1}{4}\right)^n = \frac{\frac{1}{2}}{1 - \frac{1}{2}} - \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{1}{2-1} - \frac{1}{4-1} = 1 - \frac{1}{3} = \frac{2}{3}∑n=1∞4nSn=∑n=1∞4n2n−1=∑n=1∞(21)n−∑n=1∞(41)n=1−2121−1−4141=2−11−4−11=1−31=32