How do you evaluate #\sum _ { j = 1} ^ { 3} 30( 1/3 )^j#?

1 Answer
SCooke
Dec 4, 2017

#30[13/27] = 390/27# or #14 4/9#

Explanation:

The notation means "the sum of the expression evaluated for values of 'j' from 1 to 3". The summation interval and step size can be varied, but are usually taken to be integer steps of the range.

That means the expression can be expanded into three expressions of values j = 1, 2, 3 respectively, and then added together.
#._(j = 1)^3∑30(1/3)^j# = #30(1/3)^1 + 30(1/3)^2 + 30(1/3)^3#

You may also see that algebraically this is
#30[(1/3)^1 + (1/3)^2 + (1/3)^3]# =

#30[1/3 + 1/9 + 1/27]# = #30[9/27 + 3/27 + 1/27]#

#30[13/27] = 390/27# or #14 12/27 = 14 4/9#

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