Sequences & Series

Edexcel IAL WMA12/01/P2/Jan 2020/Q8 (Arithmetic Series Proof, Sigma)

• An arithmetic series has first term a and common difference d.​​ 

Prove that the sum to n terms of this series is​​ 

n22a+n-1d 

(3)

• A sequence​​ u1 , u2 , u3 ,… is given by​​ 

un = 5n + 3–1n

Find the value of​​ 

• u5​​ 

(1)​​ 

•  ∑n=159un

(3)

SOLUTION​​ 

i- To prove the sum to n terms of arithmetic series, let us find first three terms and last 2 terms.

u1=a+n-1d=a+1-1d=a

u2=a+2-1d=a+d

u3=a+3-1d=a+2d

un-1=a+n-1-1d=a+n-2d

un=a+n-1d=a+n-1d

Hence, the series comes out to be​​ 

Sn=a+a+d+a+2d+… a+n-2d+a+n-1d           eqn (1)

Writing the reversed sum .​​ 

Sn=a+n-1d+a+n-2d+… a+2d+a+d+a          eqn (2)

Adding both eqations.​​ 

2×Sn=n[2a+n-1d]

Sn=n2[2a+n-1d]

ii- a- Putting​​ n=5​​ in the given formula.​​ 

un = 5n + 3–1n

u5=55+315

u5=25-3

u5=22

b-​​ 

∑n=159un=∑n=1595n + 3–1n

∑n=159un=∑n=1595n+∑n=1593–1n

For​​ ∑n=1595n, the series we get are​​ 

u1=51=5

u2=52=10

u3=53=15

u4=54=20

u5=55=25

This means​​ ∑n=1595n=5+10+15+20+ …​​ gives an arithmetic series with a common difference of 5. The sum of the first 59 terms of this arithmetic series can be found by the summation formula.

Using the first formula from the flashcard.​​ 

Sn=n22a+(n-1d]

S59=592[25+(59-1)×5]

S59=8850

For​​ ∑n=1593–1n, the series we get are​​ 

u1=3–1n=3-11=3-1=-3

u2=3–1n=3-12=31=3

u3=3–1n=3-13=3-1=-3

u4=3–1n=3-14=31=3

This means​​ ∑n=1593–1n=-3+3-3+3-3…​​ gives an alternating series of order 2. Since there are 59 terms,​​ which is an odd number, the sum will be the same as the first term that is -3. In other words, the positive and negative terms cancel each other out, leaving only the first term, which is -3.

So the final answer is -3.

Hence,​​ 

∑n=159un=∑n=1595n+∑n=1593–1n

∑n=159un=S59+∑n=1593–1n

∑n=159un=8850+-3

∑n=159un=8847