What is the sum #1/2+1/6+1/12+1/20+...+1/(a^2-a)# ?
1 Answer
Aug 12, 2017
Explanation:
#1/2+1/6+1/12+1/20+...+1/(a^2-a)#
#=sum_(k=2)^a 1/(k^2-k)#
#=sum_(k=2)^a 1/(k(k-1))#
#=sum_(k=2)^a (1/(k-1)-1/k)#
#=sum_(k=2)^a 1/(k-1)- sum_(k=2)^a 1/k#
#=1+color(red)(cancel(color(black)(sum_(k=2)^(a-1) 1/k))) - color(red)(cancel(color(black)(sum_(k=2)^(a-1) 1/k))) - 1/a#
#=1-1/a#
