Posted by Sarah on Friday, April 16, 2010 at 6:52pm.
Use the following formula for the sum of the cubes of the first integers to evaluate the limit in part (a).
1**3+2**3+...+n**3=((n(n+1))/2)**2
(a)lim n approaches infinity and the sum of n (top) and i=1 (bottom) with (3i/n)^3*(3/n)
I don't know how to solve this, can you help me out? Thank you.

Do not understand  Damon, Friday, April 16, 2010 at 7:10pm
Please use ^ to denote exponent
I can not figure out what your exponent really is in the summation and what you want the limit of.

Calculus (integrals)  Damon, Friday, April 16, 2010 at 7:15pm
)lim n approaches infinity and the sum of n (top) and i=1 (bottom) with (3i/n)^3*(3/n)
sum from i = 1 to i =n of what?
(3i/n)^3*(3/n)what is this?
sum from i = 1 to oo of
(3i/n)^ [3*(3/n) ]
is
sum from i = 1 to oo of
(3i/n)^ [9/n]
is that what you mean? I think not but I have no idea.

series  MathMate, Friday, April 16, 2010 at 7:46pm
If I understand correctly, the problem is to find the limit of the sum for i=1 to ∞ for the expression:
(3i/n)³(3/n)
=3³ i³ (3/n^4)
=(3/n)^4 (i)³
=(3/n)^4 ( i(i+1)/2 )²
=3^4/2^sup2; ( (i^4+2i³+i²)/n^4 )
=3^4/2^sup2; (i^4/n^4 + 2i³/n^4 + i²/n^4;)
If we take the limit as i>∞ we end up with the sum of
3^4/2^sup2;
Check my thinking.

Calculus (integrals)  Naumair, Saturday, April 17, 2010 at 3:38pm
20.25
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