Find the limit to infinity of the following equation...

Question

anonymous · Answer

$$\left[ 18-(4/3)((n)(n+1)(2n+1)/n ^{3}) \right]$$

anonymous · Answer

I kinda screwed that up. n^3 is in the denominator of the n(n+1)(2n+1)

experimentx · Answer

-8/3??

anonymous · Answer

how did you get that?

anonymous · Answer

The answer should be 46/3

turingtest · Answer

$$\lim_{n\to\infty}18-\frac43\cdot{n(n+1)(2n+1)\over n^3}$$??

experimentx · Answer

i guess 18 is not in fraction

turingtest · Answer

yeah I think I posted it right, this seems to give the right answer

anonymous · Answer

yeah that's the equation

experimentx · Answer

18 - 4/3*2 = ans

anonymous · Answer

why do you multiply -4/3 by 2?

turingtest · Answer

just look at the$${n(n+1)(2n+1)\over n^3}$$part...
distribute the top out....

turingtest · Answer

$${2n^3+3n^2+n\over n^3}$$divide top and bottom by n^3$$2+\frac3n+\frac1{n^2}$$now take the limit and you can see all this stuff becomes 2

anonymous · Answer

ooooooohhhh! thank you! I get it now! because when you distribute the top you get 2x^3 as the leading coefficient so yeah okay thanks

turingtest · Answer

the faster way to see ti is no notice that only the highest power of n will remain, so$$n\cdot n\cdot 2n=2n^3$$the n^3 on the bottom will cancel, giving two

the rest of the terms in the numerator are of degree less than 3, so they are "junk" terms that will disappear.
that is how experimentX probably saw it so fast

turingtest · Answer

you're welcome :)