Question Number 10169 by Joel575 last updated on 28/Jan/17

((2013)/1) + ((2013)/(1+2)) + ((2013)/(1+2+3)) + ... + ((2013)/(1+2+3+...+2012)) = ?

Answered by prakash jain last updated on 29/Jan/17

Σ_(i=1) ^(2012)  ((2013)/(Σ_(j=1) ^i j)) =  2013Σ_(i=1) ^(2012) (1/(i(i+1)/2))  =2013Σ_(i=1) ^(2012) (2/(i(i+1)))=2013×2Σ_(i=1) ^(2012) ((1/i)−(1/(i+1)))  =4026[(1/1)−(1/2)+(1/2)−(1/3)+.....−(1/(2012))+(1/(2012))−(1/(2013))]  =4026[1−(1/(2013))]=((4026×2012)/(2013))=2×2012

Commented bymrW1 last updated on 28/Jan/17

great technique!  please check the last step, should  it not be 2×2012?

Commented byprakash jain last updated on 29/Jan/17

You are correct. Fixed it.