Question

Need help with Exam 3 - Problem 3a)

Answer 1

This is Exam 3, problem 3a). I was completely stumped on this one. In the solution, how does one "interpret" the sum \[\sum_{i=1}^{n}(1 + \frac{ 3i }{ n })^{2}\frac{ 3 }{ n }\] as a right Riemann sum in the interval [0,3] cut into n parts?

Answer 2

It is a sum of all expression with in form such as presented for all possible values of "i" from 1 to "n". Thus it is in form of :
\[(3/n)*{[(1+3*1/n)^2]+[(1+3*2/n)^2]* ... + [(1+3*(n-1)/n)^2] +[(1+3*n/n)^2]}\]

Answer 3

(1+3i/n)^2 = 1 + 6i/n + 9i^2/n^2
3/n can be factored out of the sum and the sum can be written as sums with similar factoring:
3/n* [sum(1) + (6/n)*sum(i) + 9/n^2*sum(i^2)]
sum from 1 to n of 1 = n
sum from 1 to n of i = n(n+1)/2
sum from 1 to n of i^2 = n(n+1)(2n+1)/6
So...