Question

A= \(\dfrac{1}{2}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+....+\(\dfrac{1}{39800}\)

Answer 1

\(A=\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{39800}\)

\(=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{199\times200}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{199}-\dfrac{1}{200}\)

\(=1-\dfrac{1}{200}=\dfrac{199}{200}\)

Answer 2

\(A=\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{39800}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{199}-\dfrac{1}{200}\)

\(=\dfrac{199}{200}\)

Answer 3

199/200

Answer 4

A=1/2+1/6+...+1/39800

=1/1×2+1/2×3+...+1/199×200=1/1×2+1/2×3+...+1/199×200

=1−1/2+1/2−1/3+...+1/199−1/200=1−1/2+1/2−1/3+...+1/199−1/200

=1−1/200=199/200

Answer 5

199/200