Question

\(\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+...+59}\)

Answer 1

A = \(\dfrac{1}{1+2+3}\) + \(\dfrac{1}{1+2+3+4}\) +......+\(\dfrac{1}{1+2+3+4+....+59}\)

A = \(\dfrac{1}{(3+1).3:2}\) + \(\dfrac{1}{(4+1).4:2}\)+......+\(\dfrac{1}{(59+1).59:2}\)

A = \(\dfrac{2}{3.4}\) + \(\dfrac{2}{4.5}\) +.....+ \(\dfrac{2}{59.60}\)

A = 2.(\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{59.60}\))

A = 2. ( \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +....+ \(\dfrac{1}{59}\) - \(\dfrac{1}{60}\))

A = 2. ( \(\dfrac{1}{3}\) - \(\dfrac{1}{60}\))

A = 2. \(\dfrac{19}{60}\)

A = \(\dfrac{19}{30}\)