calculus ..... I know how to do all 3 of theM EXCEPT THE FIRST ONE.. screen shot attached
\[\sum \limits_{i=1}^n \dfrac{8i+7}{n^2}\] you can pull out n^2 because it got nothing to do with the index variable "i"
i have a previous one that i did
\[\sum \limits_{i=1}^n \dfrac{8i+7}{n^2} = \frac{1}{n^2} \sum \limits_{i=1}^n (8i+7)\]
I don't get what it's asking though?
its just asking u work that sum and express it in terms of n
thats should be the same exact answer as the one i had then? did you see the example that i got wrong?
you should get answer in terms of "n"
because the sum form i=1 to n depends on "n" right ?
yes it does
\[\begin{align} \sum \limits_{i=1}^n \dfrac{8i+7}{n^2} &= \frac{1}{n^2} \sum \limits_{i=1}^n (8i+7)\\~\\ &= \frac{1}{n^2}\left[ \sum \limits_{i=1}^n 8i+ \sum \limits_{i=1}^n7\right]\\~\\ &= \frac{1}{n^2}\left[ 8\sum \limits_{i=1}^n i+ \sum \limits_{i=1}^n7\right]\\~\\ \end{align}\]
whats the sum of first "n" natural numbers ?
what do you get when u add 7 for n times ?
\[\begin{align} \sum \limits_{i=1}^n \dfrac{8i+7}{n^2} &= \frac{1}{n^2} \sum \limits_{i=1}^n (8i+7)\\~\\ &= \frac{1}{n^2}\left[ \sum \limits_{i=1}^n 8i+ \sum \limits_{i=1}^n7\right]\\~\\ &= \frac{1}{n^2}\left[ 8\sum \limits_{i=1}^n i+ \sum \limits_{i=1}^n7\right]\\~\\ &= \frac{1}{n^2}\left[ 8\frac{n(n+1)}{2}+ 7n\right]\\~\\ \end{align}\]
simplify
I tried writing that as well, but it's just numbers no "N" expression in the answer
do u get what i'm saying?
the system doesn't want any "n" in the asnwer
S(n) depends on "n" here, so i don't know how u gona fool the system
How do u know how to put it in wolfram i tried using it once, and it gave me wrong answers, or i don't know how to use it in other words
wolfram understands latex
below also works : http://www.wolframalpha.com/input/?i=sum+%288i%2B7%29%2F%28n%5E2%29%2C+i%3D1+to+n
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