What is #1/1 + 1/(1+2) + 1/(1+2+3) + ....... + 1/(1+2+3 ...... +2015)# ?

Question

I know that the last one is 2/2016*2015 but I cannot work out the solution.

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Answer 1 · 2017-12-23T17:15:02

#"The Reqd Sum="2015/1008#.

We will solve the following more General Question :

We will find the Sum :

# s_n=1/1+1/(1+2)+1/(1+2+3)+...+1/(1+2+...+n)#.

Observe that, the general #n^(th)" term, i.e., "t_n,# is given by,

#t_n=1/(1+2+...+n)=1/(sum_(j=1)^(j=n)j)=1/{1/2n(n+1)}, or,#

# t_n=2/(n(n+1))=2*1/(n(n+1))=2{((n+1)-n)/(n(n+1))}#.

#:. t_n=2{(n+1)/(n(n+1))-n/(n(n+1))}#,

# rArr t_n=2{1/n-1/(n+1)}#.

#:. s_n=sum_(j=1)^(j=n)t_j#,

#=2{(1/1-cancel(1/2))+(cancel(1/2)-cancel(1/3))+(cancel(1/3)-cancel(1/4))+...+(cancel(1/(n-1))-cancel(1/n))+(cancel(1/n)-1/(n+1))}#,

# rArr s_n=2{1-1/(n+1)}=2*n/(n+1)#.

Clearly, #"the Reqd Sum="s_2015=2*2015/2016=2015/1008#.