1. Method of differences

2. FM Series: method of differences
KUS objectives
BAT use the method of differences to find the sum of a finite series
Starter: find the sums of the following series
1 18 (99−4𝑛)
1098
( 3 𝑛−1 )
1 𝑛 𝑟 2 +2𝑟+3 = 1 6 𝑛( 2𝑛 2 +9𝑛+25)
Show that
= 1 𝑛 𝑟 𝑛 𝑟 +3𝑛= 1 6 𝑛 𝑛+1 2𝑛+1 +2× 1 2 𝑛 𝑛+1 +3𝑛= …..

3. Notes If the general term, 𝑢 𝑟 , of a series can be expressed in the form
𝑓 𝑟 −𝑓(𝑟+1)
Then the sum of the series 1 𝑛 𝑢 𝑟 = 1 𝑛 𝑓 𝑟 −𝑓(𝑟+1)
So 𝑢 1 =𝑓 1 −𝑓 2
So 𝑢 2 =𝑓 2 −𝑓 3
So 𝑢 3 =𝑓 3 −𝑓(2) …
So 𝑢 𝑛 =𝑓 𝑛 −𝑓(𝑛+1)
What happens when we sum all these terms?
Lots of cancelling!
1 𝑛 𝑢 𝑟 =𝑓 1 −𝑓(𝑛+1)

4. WB1 a) Show that 4 𝑟 3 = 𝑟 2 (𝑟+1) 2 − (𝑟−1) 2 𝑟 2
b) Hence, prove by a method of differences, that
1 𝑛 𝑟 3 = 1 4 𝑛 2 (𝑛+1) 2
a) RHS 𝑟 2 (𝑟+1) 2 − 𝑟−1 2 𝑟 2 = 𝑟 2 𝑟 2 +2𝑟+1 − 𝑟 2 𝑟 2 −2𝑟+1
= 𝑟 4 +2 𝑟 3 + 𝑟 2 − 𝑟 4 +2 𝑟 3 − 𝑟 2 = 4𝑟 QED
1 𝑛 𝑟 2 (𝑟+1) 2 − 𝑟−1 2 𝑟 2
b) Consider
So r= 𝑢 1 = −
So r= 𝑢 2 = −
So r= 𝑢 3 = − …
So 𝑢 𝑛 = 𝑛 2 (𝑛+1) 2 − (𝑛−1) 2 𝑛 2
Cancel and
Sum together
4 1 𝑛 𝑟 3 = 𝑛 2 (𝑛+1) 2 − = 𝑛 2 (𝑛+1) 2
1 𝑛 𝑟 3 = 𝑛 2 (𝑛+1) 2

5. WB2 a) verify that 1 𝑟 𝑟+1 = 1 𝑟 − 1 𝑟+1
b) Hence, use a method of differences to find
1 𝑛 1 𝑟 𝑟+1
a) RHS 𝑟 − 1 𝑟+1 = 𝑟+1−𝑟 𝑟 𝑟+1 = 1 𝑟(𝑟+1) QED
1 𝑛 1 𝑟 − 1 𝑟+1
b) Consider
So r= 𝑢 1 = 1 1 − 1 2
So r= 𝑢 2 = 1 2 − 1 3
So r= 𝑢 3 = 1 3 − 1 4 …
So 𝑢 𝑛 = 1 𝑛 − 1 𝑛+1
Cancel and
Sum together
1 𝑛 1 𝑟 𝑟+1 =1− 1 𝑛+1 = 𝑛 𝑛+1

6. WB3 use a method of differences to find
1 𝑛 1 4 𝑟 2 −1
WB3 use a method of differences to find
a) First split into partial fractions 𝑟 2 −1 = 1 2𝑟+1 2𝑟−1 = 𝐴 2𝑟+1 + 𝐵 2𝑟−1
= −1/2 2𝑟+1 + 1/2 2𝑟−1
1 𝑛 𝑟−1 − 1 2𝑟+1
b) Consider
= 𝑟−1 − 1 2𝑟+1
So r= 𝑢 1 = 1 1 − 1 3
So r= 𝑢 2 = 1 3 − 1 5
So r= 𝑢 3 = 1 5 − 1 7 …
So 𝑢 𝑛 = 1 2𝑛−1 − 1 2𝑛+1
Cancel and
Sum together
𝑛 1 2𝑟−1 − 1 2𝑟+1 = − 1 2𝑛+1 =…= 𝑛 2𝑛+1

7. WB4 a) use a method of differences to show that
where a, b are constants to be found
b) Find 𝑟+1 𝑟 to 5 decimal places
1 𝑛 2 𝑟+1 𝑟+3 = 𝑛 𝑎𝑛+𝑏 6 𝑛+2 𝑛+3
a) First split into partial fractions 𝑟+1 𝑟+3 = 𝐴 𝑟+1 + 𝐵 𝑟+3
= 1 𝑟+1 − 1 𝑟+3
1 𝑛 1 𝑟+1 − 1 𝑟+3
Consider
So r= 𝑢 1 = 1 2 − 1 4
So r= 𝑢 2 = 1 3 − 1 5
So r= 𝑢 3 = 1 4 − 1 6 …
So r=n-1 𝑢 𝑛−1 = 1 𝑛 − 1 𝑛+2
So 𝑢 𝑛 = 1 𝑛+1 − 1 𝑛+3
Cancel and
Sum together
𝑛 1 𝑟+1 − 1 𝑟+3 = − 1 𝑛+2 − 1 𝑛+3 =…= 𝑛(5𝑛+13) 6(𝑛+2)(𝑛+3)

8. WB4 (cont) a) use a method of differences to show that
where a, b are constants to be found
b) Find 𝑟+1 𝑟 𝑡𝑜 5 𝑑𝑒𝑐𝑖𝑚𝑎𝑙 𝑝𝑙𝑎𝑐𝑒𝑠
1 𝑛 2 𝑟+1 𝑟+3 = 𝑛 𝑎𝑛+𝑏 6 𝑛+2 𝑛+3
𝑟+1 𝑟+3 = 𝑟+1 𝑟+3 − 𝑟+1 𝑟+3 b)
= 30(150+13) − = = (5𝑑𝑝)
NOW DO EX 2A

9. One thing to improve is –
KUS objectives
BAT use the method of differences to find the sum of a finite series
self-assess
One thing learned is –
One thing to improve is –

10. END