Question

The sum of infinite series 4+8/3+12/3^2+16/3^3....

Answer 1

Given : 4  + 8/3  + 12/3²  + 16/3³ + ...............................

To Find : sum of infinite series

Solution:

4  + 8/3  + 12/3²  + 16/3³ + ...............................

4 + (4/3 + 4/3)  + (4/3² + 4/3² + 4/3²)  +(4/3³ + 4/3³+ 4/3³ + 4/3³)

Hence we can convert this infinite series in a lot of infinite series

with r = 1/3  and  

a = 4  ,  4/3  ,  4/3² , 4/3³  and so on

Sum of 1st series   using Sn = a/(1 - r)

= 4/(1 - 1/3)   = 6

Sum of 2nd Series  

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

= 2

Sum of 3rd Series  

= (4/3²)/(1 - 1/3)

= 2/3

Hence it forms another infinite GP

with a = 6  r = 1/3

Sum =  6 /(1 - 1/3)   =  9

9 is Sum of infinite series 4  + 8/3  + 12/3²  + 16/3³ + ...............................

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