Find the sum of the first ten terms of the sequence \(\left\{\left(\frac{1}{2}\right)^{n-1}\right\}\)
\(S=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+\frac{1}{2^9}\)
\(\frac{1}{2}S=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots+\frac{1}{2^{10}}\)
\(S-\frac{1}{2}S=1-\frac{1}{2^{10}}\)
\(\frac{1}{2}S=1-\frac{1}{2^{10}}\)
\(S=2-\frac{1}{2^{9}}=\frac{1023}{512}\approx 1.998\)
Find the sum of the first ten terms of the sequence \(\left\{3\left(-\frac{1}{3}\right)^{n-1}\right\}\).
\(S=3-1+\frac{1}{3}-\frac{1}{9}+\cdots+\left(-\frac{1}{3}\right)^8\)
\(-\frac{1}{3}S=-1+\frac{1}{3}-\frac{1}{9}+\frac{1}{27}+\cdots+\left(-\frac{1}{3}\right)^9\)
\(S-\left(-\frac{1}{3}S\right)=3-\left(-\frac{1}{3}\right)^9=3+\left(\frac{1}{3}\right)^9\)
\(\frac{4}{3}S=3+\left(\frac{1}{3}\right)^9\)
\(S=\frac{9}{4}+\frac{1}{4(3^8)}=\frac{29525}{13122}\approx 2.25\)