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Mathematics

Find the sum of G.P. :

1−12+14−18+.......1 - \dfrac{1}{2} + \dfrac{1}{4} - \dfrac{1}{8} + ……. to 9 terms

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Answer

Common ratio (r) = −121=−12\dfrac{-\dfrac{1}{2}}{1} = -\dfrac{1}{2}

S=a(1−rn)(1−r)..........(As∣r∣<1)=1[1−(−12)9]1−(−12)=[1+(129)]1+12=(1+129)32=23(1+129)S = \dfrac{a(1 - r^n)}{(1 - r)} ……….(As |r| \lt 1) \\[1em] = \dfrac{1\Big[1 - \Big(-\dfrac{1}{2}\Big)^9\Big]}{1 - \Big(-\dfrac{1}{2}\Big)} \\[1em] = \dfrac{\Big[1 +\Big(\dfrac{1}{2^9}\Big)\Big]}{1 + \dfrac{1}{2}} \\[1em] = \dfrac{\Big(1 + \dfrac{1}{2^9}\Big)}{\dfrac{3}{2}} \\[1em] = \dfrac{2}{3}\Big(1 + \dfrac{1}{2^9}\Big)

Hence, sum = 23(1+129)\dfrac{2}{3}\Big(1 + \dfrac{1}{2^9}\Big).

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