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.
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.
4 answers
)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.
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.
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.
(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.
20.25
2 answers