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Method of differences.

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Presentation on theme: "Method of differences."— Presentation transcript:

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


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