Given sequence is :
3,-1,\(\frac{1}{3}\),\(\frac{-1}{9}\),....
∴ a1 = 3, a2 = -1, a3 = \(\frac{1}{3}\), a4 = \(\frac{-1}{9}\),....
Common ration = r = \(\frac{a_4}{a_3}\) = \(\frac{a_3}{a_2}\) = \(\frac{a_2}{a_1}\) = \(\frac{-1}{3}\) < 1
nth term of given sequence is,
an = arn-1
= 3 x (\(\frac{1}{3}\))n-1
= (-1)n-1 x 3 x \(\frac{1}{3^{n-1}}\)
= \(\frac{(-1)^{n-1}}{3^{n-2}}\)
Sum upto 13 terms = S13 = \(\frac{a(1-r^{13})}{1-r}\)
= \(\frac{3(1-(\frac{-1}{3})^{13})}{1-(\frac{-1}{3})}\)
= \(\frac{3(1+\frac{1}{3^{13}})}{1+\frac{1}{3}}\)
= \(\frac{3(1+\frac{1}{3^{13}})}{\frac{4}{3}}\)
= \(\frac{9}{4}\)(1 + \(\frac{1}{3^{13}}\))