(1) 13n(2n−1)=2n−13n\frac{1}{3^n}(2^n-1)=\frac{2^n-1}{3^n}3n1(2n−1)=3n2n−1 (2) ∑n=1n2n−13n=∑n=1n(23)n−∑n=1n(13)n=23{1−(23)n}1−23−13{1−(13)n}1−13=2{1−(23)n}−12{1−(13)n}\sum_{n=1}^{n}\frac{2^n-1}{3^n}=\sum_{n=1}^{n}\left(\frac{2}{3}\right)^n-\sum_{n=1}^{n}\left(\frac{1}{3}\right)^n=\frac{\dfrac{2}{3}\left\{1-\left(\dfrac{2}{3}\right)^n\right\}}{1-\dfrac{2}{3}}-\frac{\dfrac{1}{3}\left\{1-\left(\dfrac{1}{3}\right)^n\right\}}{1-\dfrac{1}{3}}=2\left\{1-\left(\frac{2}{3}\right)^n\right\}-\frac{1}{2}\left\{1-\left(\frac{1}{3}\right)^n\right\}∑n=1n3n2n−1=∑n=1n(32)n−∑n=1n(31)n=1−3232{1−(32)n}−1−3131{1−(31)n}=2{1−(32)n}−21{1−(31)n} (3) S=limn→∞[2{1−(23)n}−12{1−(13)n}]=2(1−0)−12(1−0)=32S=\lim_{n\to\infty}\left[2\left\{1-\left(\frac{2}{3}\right)^n\right\}-\frac{1}{2}\left\{1-\left(\frac{1}{3}\right)^n\right\}\right]=2(1-0)-\frac{1}{2}(1-0)=\frac{3}{2}S=limn→∞[2{1−(32)n}−21{1−(31)n}]=2(1−0)−21(1−0)=23