Question

Transform the following summation by making the change of variable j = i - 9. When i = 10, then j = . So when i and ((n - i))^2 are expressed in terms of j, the results are i = and ((n - i))^2 = . Thus, (sum_{i=9}^{n-1} frac{i}{(n-i)^2} ) = (sum_{j=0}^{n-10} )

          Transform the following summation by making the change of variable j = i - 9.
When i = 10, then j =  . So when i and ((n - i))^2 are expressed in terms of j, the results are
i =  and ((n - i))^2 =  . Thus,
(sum_{i=9}^{n-1} frac{i}{(n-i)^2} ) = (sum_{j=0}^{n-10} )

$Transform the following summation by making the change of variable j = i - 9. When i = 10, then j = . So when i and ((n - i))^2 are expressed in terms of j, the results are i = and ((n - i))^2 = . Thus, (sumi=9^n-1 fraci(n-i)^2 ) = (sumj=0^n-10 )$

Added by Bryce M.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert Brittany Stefanilo

Step 1

Then, $i = j + 9$. We also need to change the limits of the summation. When $i = 1$, we have $j = 1 - 9 = -8$. When $i = n$, we have $j = n - 9$. So, the new summation is: $$\sum_{j=-8}^{n-9} (n - (j + 9))^2$$ Now, let's simplify the expression inside the Show more…

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Transform the following summation by making the change of variable j = i - 9. When i = 10, then j = 1. So when n - 7 and (n - 7)^2 are expressed in terms of j, the results are n - 8 and (n - 1)^2, respectively.