How do you solve this sigma problem?

sum3(5)n-1

There is an 8 above the sigma and n=1

2 Answers
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

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