liya2722
Answered

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Consider Sn = 1/2 + 1/4 + 1/8 + 1/16 +...+ 1/2^n a Find S1, S2, S3, S4 and S5 in fractional form b From A guess the formula for Sn c Find Sn using Sn= u1(1-r^n)/1-r d Comment on S, as n gets very large. e What is the relationship between the given diagram and d? ​

Sagot :

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.

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