What is the sum #1/2+1/6+1/12+1/20+...+1/(a^2-a)# ?

1 Answer
Aug 12, 2017

Answer:

#1/2+1/6+1/12+1/20+...+1/(a^2-a)=1-1/a#

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#