# How do you solve this sigma problem?

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#sum# 3(5)n-1

There is an 8 above the sigma and n=1

There is an 8 above the sigma and n=1

##### 2 Answers

#### Explanation:

Not sure of your notation here. I am assuming it is:

We can recognise

Where:

The sum of a geometric series is given as:

So plugging in the known values:

#### Explanation:

#"assuming "sum_(n=1)^8 3(5)n-1#

#"using the "color(blue)"summation blocks"#

#•color(white)(x)sum_(r=1)^n 1=n#

#•color(white)(x)sum_(r=1)^n r=1/2n(n+1)#

#"the question simplifies to"#

#sum_(n=1)^8 15n-1#

#=15sum_(n=1)^8 n-sum_(n=1)^8 1#

#=15/2xx(8xx9)-8#

#=540-8=532#