**Given sequence is :**

3,-1,\(\frac{1}{3}\),\(\frac{-1}{9}\),....

∴ a_{1} = 3, a_{2} = -1, a_{3} = \(\frac{1}{3}\), a_{4} = \(\frac{-1}{9}\),....

Common ration = r = \(\frac{a_4}{a_3}\) = \(\frac{a_3}{a_2}\) = \(\frac{a_2}{a_1}\) = \(\frac{-1}{3}\) < 1

**n**^{th} term of given sequence is,

a_{n} = ar^{n-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 = S_{13} = \(\frac{a(1-r^{13})}{1-r}\)

= \(\frac{3(1-(\frac{-1}{3})^{13})}{1-(\frac{-1}{3})}\)

= \(\frac{3(1+\frac{1}{3^{13}})}{1+\frac{1}{3}}\)

= \(\frac{3(1+\frac{1}{3^{13}})}{\frac{4}{3}}\)

= \(\frac{9}{4}\)(1 + \(\frac{1}{3^{13}}\))