Question

a.\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+......+\frac{1}{512}+\frac{1}{1024}\)

b.\(\frac{1}{8}+\frac{1}{48}+\frac{1}{80}+.....+\frac{1}{10200}\)

Answer 1

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Answer 2

a/ \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)

= \(\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+...+\left(\frac{1}{256}-\frac{1}{512}\right)+\left(\frac{1}{512}-\frac{1}{1024}\right)\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{256}-\frac{1}{512}+\frac{1}{512}-\frac{1}{1024}\)

= \(1-\frac{1}{1024}\)

= \(\frac{1023}{1024}\)

b/ \(\frac{1}{8}+\frac{1}{48}+\frac{1}{80}+...+\frac{1}{10200}\)

= \(\frac{1}{8}+\frac{1}{6\times8}+\frac{1}{8\times10}+...+\frac{1}{100\times102}\)

= \(\frac{1}{8}+\frac{1}{2}\times\left(\frac{2}{6\times8}+\frac{2}{8\times10}+...+\frac{2}{100\times102}\right)\)

= \(\frac{1}{8}+\frac{1}{2}\times\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{102}\right)\)

= \(\frac{1}{8}+\frac{1}{2}\times\left(\frac{1}{6}-\frac{1}{102}\right)\)

= \(\frac{1}{8}+\frac{1}{2}\times\frac{8}{51}\)

= \(\frac{1}{8}+\frac{4}{51}\)

= \(\frac{83}{408}\)

Answer 3

a.1/2+1/4+1/8+...+1/512+1/1024

=1-1/2+1/2-1/4+1/4-1/8+...+1/256-1/512+1/512-1/1024

=1-1/1024

=1023/1024

b.