$\displaystyle \sum_{i=0}^n\frac{4}{n}*({1+(i/4n)})^4\$ as n goes infinite
Last edited by chialin4; Feb 12th 2010 at 06:03 AM.
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Note that the sum is the Riemann Sum of f(x) = (1+x/16)^4 on the interval 0 to 4 So the sum = integral[(1+x/16)^4dx] from 0 to 4
Originally Posted by Calculus26 Note that the sum is the Riemann Sum of f(x) = (1+x/16)^4 on the interval 0 to 4 So the sum = integral[(1+x/16)^4dx] from 0 to 4 i think the interval is 1 to 5?? is it right??
No it is 0 to 4 see the attachment where I use mathcad to evaluate the sum and the integral to verify my results. Why do you think the interval would be 1 to 5 ?
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