How do you solve this sigma problem?

#sum#3(5)n-1

There is an 8 above the sigma and n=1

2 Answers
Somebody N.
May 11, 2018

#color(blue)(292968)#

Explanation:

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

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

We can recognise #3*(5)^(n-1)# as being the nth term of a geometric series:

#ar^(n-1)#

Where:

#bba# is the first term, #bbr# is the common ratio and #bbn# is the nth term:

The sum of a geometric series is given as:

#a((1-r^n)/(1-r))#

So plugging in the known values:

#3((1-5^8)/(1-5))=3((-390624)/-4)=3(97656)=292968#

#sum_(n=1)^(8)3*(5)^(n-1)=292968#

Jim G.
May 11, 2018

#532#

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#

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