Summation of series(method of differences),When we use #∑f(r)-f(r+1#),how can we know it is #f(1)-f(n+1)#, but not #f(n)-f(1+1)#? Also, how can we know it is #f(n+1)-f(1)# but not #f(1+1)-f(n)# when we use #∑f(r+1)-f(r)#?

Question

Liam Brennan · Accepted Answer

# sum_(r=1)^n f(r)-f(r+1) = f(1)-f(n+1)#
# sum_(r=1)^n f(r+1)-f(r) = f(n+1) -f(1)#

Write out the terms and see what happens: Consider: # S_1 = sum_(r=1)^n f(r)-f(r+1) # # " " = {f(1)-f(2)} + # # " " {f(2)-f(3)} + # # " " {f(3)-f(4)} + # # " " {f(4)-f(5)} + ... + # # " " {f(n-1)-f(n)} + # # " " {f(n)-f(n+1)} # # " " = {f(1)-cancel(color(purple)f(2))} + # # " " {cancel(color(purple)f(2))-cancel(color(red)f(3))} + # # " " {cancel(color(red)f(3))-cancel(color(green)f(4))} + # # " " {cancel(color(green)f(4))-cancel(f(5))} + ... + # # " " {cancel(f(n-1))-cancel(color(blue)f(n))} + # # " " {cancel(color(blue)f(n))-f(n+1)} # And as shown almost all terms vanish, leaving: # S_1 = f(1)-f(n+1) # Whereas with: # S_2 = sum_(r=1)^n f(r+1)-f(r) # # " " = {f(2)-f(1)} + # # " " {f(3)-f(2)} + # # " " {f(4)-f(3)} + # # " " {f(5)-f(4)} + ... + # # " " {f(n)-f(n-1)} + # # " " {f(n+1)-f(n)} # # " " = {cancel(color(purple)f(2))-f(1)} + # # " " {cancel(color(red)f(3))-cancel(color(purple)f(2))} + # # " " {cancel(color(green)f(4))-cancel(color(red)f(3)}) + # # " " {cancel(f(5))-cancel(color(green)f(4))} + ... + # # " " {cancel(color(blue)f(n))-cancel(f(n-1))} + # # " " {f(n+1)-cancel(color(blue)f(n))} # Similarly, almost all terms vanish, leaving: # S_2 = -f(1) + f(n+1) # # " " = f(n+1) -f(1)#

Cora Alexander · Answer

Refer to the Explanation.

Let us say, #S=sum_(r=1)^n{f(r)-f(r+1)}.# To find #S,# we substitute, #r=1,2,3,...,n-1,n,# successively in #{f(r)-f(r+1)},# and get, # S={f(1)-cancelf(2)}+{cancelf(2)-cancelf(3)}+{cancelf(3)-cancelf(4)}# #+...+{cancelf(n-1)-cancelf(n)}+{cancelf(n)-f(n+1)},# # rArr S=sum_(r=1)^n{f(r)-f(r+1)}=f(1)-f(n+1).# This proves the #1^(st)# Assertion. The Proof of the #2^(nd)# Assertion can be obtained similarly. Otherwise, by what we have proved above, # sum_(r=1)^n{f(r)-f(r+1)}=f(1)-f(n+1).# Multiplying this by #-1,# we have, # -sum_(r=1)^n{f(r)-f(r+1)}=-{f(1)-f(n+1)}, i.e., # #:. sum_(r=1)^n{f(r+1)-f(r)}=f(n+1)-f(1).# Enjoy Maths.!