Question

a)1/2+1/6+1/12+........+1/9900+1/10100

b)1/2+1/4+1/8+....+1/256+1/512

Answer 1

a, \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{10100}\)

=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{100.101}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{100}-\frac{1}{101}\)

=\(1-\frac{1}{101}\)

=\(\frac{100}{101}\)

b,\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{512}\)

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

=\(1-\frac{1}{512}\)

=\(\frac{511}{512}\)

Answer 2

\(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}+\frac{1}{10100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\)

\(=1-\frac{1}{101}=\frac{100}{101}\)