# 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)#?

##### 2 Answers

# 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)#

#### Explanation:

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)#

Refer to the **Explanation.**

#### Explanation:

Let us say,

This proves the **Assertion.**

The **Proof** of the **Assertion** can be obtained **similarly.**

**Otherwise,** by what we have proved above,

Multiplying this by

**Enjoy Maths.!**