Find the sum of the first six terms in the geometric sequence 1/2, 1/8, 1/16, .... Express your answer as a common fraction.
+526 I think you missed a term, the GP should be : 1/2, 1/4, 1/8, 1/16,...
⇒ a = 1/2
r = 1/2
n = 6
\({S}_{n}={a(1-r^n)\over 1-r}\)
\(= {{1\over 2}(1- {1\over 2^6})\over {1- {1\over 2}}}\)
\(=1- {1\over 2^6}\)
\(={63 \over 64}\)
\({S}_{6}= {63\over 64}\)
+526 I think you missed a term, the GP should be : 1/2, 1/4, 1/8, 1/16,...
⇒ a = 1/2
r = 1/2
n = 6
\({S}_{n}={a(1-r^n)\over 1-r}\)
\(= {{1\over 2}(1- {1\over 2^6})\over {1- {1\over 2}}}\)
\(=1- {1\over 2^6}\)
\(={63 \over 64}\)
\({S}_{6}= {63\over 64}\)
