Question

Question about geometric progression?

1.find the sum of n terms of the G.P 3+ 3/2+ 3/4 + 3/8 +.....,show that,however great n may be,the sum cannot exceed 6.

2.Sum the series 3+ 5/2 + 7/4 + ....+ 2n+1/2^n-1 . (answer 6[1-(1/2)^n, 10- 1/2^n(4n+10) )

Please help me solve this question step by step,Thank you.

Answer

1. S = a + ar + ar^2 + ar^3 + ... + ar^n

Multiplying the above equation through by r gives:

rS = ar + ar^2 + ar^3 + ... + ar^n + ar^(n+1)

Subtracting the second equation from the first gives:

(S - rS) = a - ar^(n+1)

Factoring gives:

S(1 - r) = a(1 - r^(n+1))

Therefore, S = a(1 - r^(n+1)) / (1 - r).

As n gets very large, r^(n+1) tends to zero since the absolute value of r is less than 1.

This leaves S = a / (1 - r) as the number of terms goes to infinity.

S = 3 / (1 - 1/2)) = 3 / (1/2) = 6.

2.

Nov 20 at 5:21

Sn = 3 + 3/2 +3/4 +3/8 + .......

a = 3 r = 1/2

Sn = a(r^n -1) /( r -1 ) if r >1

= a( 1 - r^n) /(1 -r) if | r |<1

lim a( 1 - r^n) /(1 -r) | r^ n -> 0 if r < 1

n -> infinity

a /(1 - r)

= 3/( 1-1/2) = 6

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2.Sum the series 3+ 5/2 + 7/4 + ....+ 2n+1/2^n-1 . (answer 6[1-(1/2)^n, 10- 1/2^n(4n+10) )

This is aritho - geometric series

Nov 20 at 9:7