Asked by Lisa
5
Σ 4(3/4)^k
k=0
Find the sum of the series.
Round the answer to three decimal places.
Σ 4(3/4)^k
k=0
Find the sum of the series.
Round the answer to three decimal places.
Answers
Answered by
Reiny
5
Σ 4(3/4)^k
k=0
= 4(3/4)^0 + 4(3/4)^1 + 4(3/4)^2 + 4(3/4)^3 + 4(3/4)^4 + 4(3/4)^5
= 4[ 1 + (3/4) + (3/4)^2 + (3/4)^3 + (3/4)^4 + (3/4)^5 ]
The part in the brackets is a GS with a = 1 , r = 3/4 and n = 6
= 4 ( 1( (3/4)^6 - 1)/(3/4 - 1)
=4 (729/4096 - 1)/(-1/4)
= -16(-3367/4096)
= 3367/256 ---- > exact answer
= 13.152 ----> correct to 3 decimals
Σ 4(3/4)^k
k=0
= 4(3/4)^0 + 4(3/4)^1 + 4(3/4)^2 + 4(3/4)^3 + 4(3/4)^4 + 4(3/4)^5
= 4[ 1 + (3/4) + (3/4)^2 + (3/4)^3 + (3/4)^4 + (3/4)^5 ]
The part in the brackets is a GS with a = 1 , r = 3/4 and n = 6
= 4 ( 1( (3/4)^6 - 1)/(3/4 - 1)
=4 (729/4096 - 1)/(-1/4)
= -16(-3367/4096)
= 3367/256 ---- > exact answer
= 13.152 ----> correct to 3 decimals
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