S_1 =sum n=1 to 1] 4 * (1/3)^n = 4 * (1/3) = 4/3 S_2 = sum n=1 to 2] 4 * (1/3)^n

Question

S_1 =sum n=1 to 1] 4 * (1/3)^n

= 4 * (1/3)

= 4/3

S_2 = sum n=1 to 2] 4 * (1/3)^n

= 4 * (1/3) + 4 * (1/3)^2

= 4/3 + 4/9

= 16/9

S_3 =[sum n=1 to 3] 4 * (1/3)^n

= 4 * (1/3) + 4 * (1/3)^2 + 4 * (1/3)^3

= 4/3 + 4/9 + 4/27

= 52/27

S_4 =[sum n=1 to 4] 4 * (1/3)^n

= 4 * (1/3) + 4 * (1/3)^2 + 4*(1/3)^3 + 4*(1/3)^4

= 4/3 + 4/9 + 4/27 + 4/81

= 160/81

S_5= [sum n=1 to 5] 4 * (1/3)^n

= 4 * (1/3) + 4 * (1/3)^2 + 4*(1/3)^3 + 4*(1/3)^4 + 4*(1/3)^5

= 4/3 + 4/9 + 4/27 + 4/81 + 4/243

= 484 / 243

How do we know for sure that the above infinite geometric series has a sum?   What is the formula that we can use to find the sum of an infinite geometric series?

Explanation / Answer

in case of infinite series then go for common ratio = (1/3) here

sum of infinite series = a / (1-r)

= (4/3) / ( 1 - 1/3)

= 2

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