Question

B = \(\frac{1}{3}\)+ \(\frac{1}{9}\)+ \(\frac{1}{27}\) + ........................... + \(\frac{1}{6561}\)

 

Answer 1

\(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\)

\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)

\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)

\(3B-B=\left(1+\frac{1}{3}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)\)

\(2B=1-\frac{1}{3^8}\)

\(B=\frac{1-\frac{1}{3^8}}{2}\)

Answer 2

B = 1/3 + 1/9 + 1/27 + ... + 1/6561 

B = 1/3^1 + 1/3^2 + 1/3^3 + ... + 1/3^8 

3B = 1 + 1/3^1 + 1/3^2 + ... + 1/3^7 

3B - B = ( 1 + 1/3^1 +1/3^2 + ... + 1/3^7 ) - ( 1/3^1 + 1/3^2 + 1/3^3 + .... + 1/3^8 ) 

2B = 1 - 1/3^8

B = 1 - 1/3^8  / 2

Answer 3

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