Find the sum of G.P. :
1−13+132−1331 - \dfrac{1}{3} + \dfrac{1}{3^2} - \dfrac{1}{3^3}1−31+321−331 + …… to n terms
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Common ratio (r) = −131=−13\dfrac{-\dfrac{1}{3}}{1} = -\dfrac{1}{3}1−31=−31.
S=a(1−rn)(1−r)..........(As∣r∣<1)=1[1−(−13)n]1−(−13)=[1−(−13)n]1+13=[1−(−13)n]43=34[1−(−13)n].S = \dfrac{a(1 - r^n)}{(1 - r)} ……….(As |r| \lt 1) \\[1em] = \dfrac{1\Big[1 - \Big(-\dfrac{1}{3}\Big)^n\Big]}{1 - \Big(-\dfrac{1}{3}\Big)} \\[1em] = \dfrac{\Big[1 - \Big(-\dfrac{1}{3}\Big)^n\Big]}{1 +\dfrac{1}{3}} \\[1em] = \dfrac{\Big[1 - \Big(-\dfrac{1}{3}\Big)^n\Big]}{\dfrac{4}{3}} \\[1em] = \dfrac{3}{4}\Big[1 - \Big(-\dfrac{1}{3}\Big)^n\Big].S=(1−r)a(1−rn)……….(As∣r∣<1)=1−(−31)1[1−(−31)n]=1+31[1−(−31)n]=34[1−(−31)n]=43[1−(−31)n].
Hence, sum = 34[1−(−13)n].\dfrac{3}{4}\Big[1 - \Big(-\dfrac{1}{3}\Big)^n\Big].43[1−(−31)n].
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