How do you solve this sigma problem?
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
May 11, 2018
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:
May 11, 2018
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