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Calculus (integrals)

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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 - ,

    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) - ,

    )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 - ,

    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) - ,

    20.25

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