Integral by limit of sum
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HallsofIvy
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You correctly write this as \(\displaystyle \sum_{i=0}^n \left(\frac{5i}{n}+ 1\right)\frac{5}{n}= \frac{25}{n^2}\sum_{i=0}^n i+ \frac{5}{n}\sum_{i= 0}^n 1\).
So what is \(\displaystyle \sum_{i= 0}^n i\)? With n= 1 it is 1, with n= 2, it is 1+ 2= 3, with n= 3, 1+ 2+ 3= 6. In general it is S= 1+ 2+ 3+ . . . + (n-2)+ (n-1)+ n. If we turn it around it is S= n+ (n-1)+ (n-2)+ . . .+ 3+ 2+ 1. Adding each part, n+ 1, (n-1)+ 2= n+1, (n-2)+ 3= n+1, etc. So the sum, 2S is equal to n time n+ 1. \(\displaystyle \sum_{i=0}^n i= \frac{n(n+1)}{2}\). Do you see that 1(1+1)/2= 1, 2(2+1)/2= 3, and 3(3+1)/2= 6?
And, as Jomo said, adding 1 to itself n times is NOT 1! What is it? Well, 1+ 1= 2, 1+ 1+ 1= 3, 1+ 1+ 1+ 1= 4, etc. Pretty easy isn't it?
So what is \(\displaystyle \sum_{i= 0}^n i\)? With n= 1 it is 1, with n= 2, it is 1+ 2= 3, with n= 3, 1+ 2+ 3= 6. In general it is S= 1+ 2+ 3+ . . . + (n-2)+ (n-1)+ n. If we turn it around it is S= n+ (n-1)+ (n-2)+ . . .+ 3+ 2+ 1. Adding each part, n+ 1, (n-1)+ 2= n+1, (n-2)+ 3= n+1, etc. So the sum, 2S is equal to n time n+ 1. \(\displaystyle \sum_{i=0}^n i= \frac{n(n+1)}{2}\). Do you see that 1(1+1)/2= 1, 2(2+1)/2= 3, and 3(3+1)/2= 6?
And, as Jomo said, adding 1 to itself n times is NOT 1! What is it? Well, 1+ 1= 2, 1+ 1+ 1= 3, 1+ 1+ 1+ 1= 4, etc. Pretty easy isn't it?