Part (a)
S1 = 1/2 since we only have one term, i.e. we aren't really summing anything here.
S2 = (1/2) + (1/4) = (2/4) + (1/4) = (2+1)/4 = 3/4
S3 = (1/2) + (1/4) + (1/8) = (4/8) + (2/8) + (1/8) = (4+2+1)/8 = 7/8
Or,
S3 = S2 + 1/8 = (3/4) + (1/8) = (6/8) + (1/8) = (6+1)/8 = 7/8
S4 = S3 + 1/16 = (7/8) + (1/16) = (14/16) + (1/16) = (14+1)/16 = 15/16
S5 = S4 + 1/32 = (15/16) + (1/32) = (30/32) + (1/32) = (30+1)/32 = 31/32
Here's the summary:
- S1 = 1/2
- S2 = 3/4
- S3 = 7/8
- S4 = 15/16
- S5 = 31/32
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Part (b)
Through trial and error, the formula is
Sn = ( (2^n) -1 )/(2^n)
For example, if n = 4, then,
Sn = ( (2^n) -1 )/(2^n)
S4 = ( (2^4) -1 )/(2^4)
S4 = (16 - 1)/(16)
S4 = 15/16
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Part (c)
u1 = first term = 1/2
r = common ratio = 1/2
Sn = u1*( 1 - r^n)/(1 - r)
Sn = (1/2)*( 1 - (1/2)^n )/(1 - 1/2)
Sn = (1/2)*( 1 - (1/2)^n )/(1/2)
Sn = 1 - (1/2)^n
Sn = 1 - 1/(2^n)
Sn = (2^n)/(2^n) - 1/(2^n)
Sn = ( (2^n) -1 )/(2^n)
This helps confirm the formula mentioned in part b
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Part (d)
As n gets very large, Sn approaches 1
Examples:
- n = 7 leads to S7 = 0.9921875
- n = 10 leads to S10 = 0.9990234375
- n = 12 leads to S12 = 0.999755859375
- n = 20 leads to S20 = 0.999999046325684
Try out larger values of n and the Sn value should get closer to 1.
The reason why this works is because Sn = (x - 1)/x where x = 2^n
The (x-1)/x is equivalent to 1 - (1/x) and the 1/x approaches zero when x gets larger. The x gets larger as 2^n does.
The value of Sn itself will never equal 1 exactly. It simply gets closer and closer.
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Part (e)
The diagram is missing, so I don't have enough info to be able to answer this.
