:( this question is killing me Write the limit as n goes to infinity of the summation from k equals 1 of the product of the 3rd power of the quantity 2 plus 5 times k over n and 5 over n as a definite integral.
@amistre64 @welshfella any ideas?
not really sure how to go about this one.
@Preetha @nincompoop @Hero @Luigi0210 @Nnesha
what is the limit of 5/n as n to infinity?
the limit of a product is the product of the limits, if i recall correctly .. but you want the summation limit
0?
yes, now if there is some useful way to rewrite the argument ...
(a+b)^3 = 1a^3b^0 +3a^2b^1 +3a^1b^2 +1a^0b^3
im sorry i dont follow
im wondering if we expand the ^3 part, if we cant see this thing easier, or make it funner to play with.
i dont think that would make a difference
40/n + 300k/(n^2) +750k^2/(n^3) + 625k^3/(n^4) youre prolly right
well, by simply putting in a large number for n http://www.wolframalpha.com/input/?i=sum%28n%3D1+to+100000%29+of+5%282%2Bn%285%2F100000%29%29^3%2F100000
don't really need to find the limit though just need to know how to write it as an integral
do you know anyone that could solve this , im sort of new to openstudy don't know anyone :(
oh, i gave up trying to read the whole question when the numbers were written out in words ...
so this is a reimann sum thing
yes can you open this picture ?
\[\Large\sum_{k=a}^{b}f(a+i\frac{b-a}{n})\frac{b-a}{n}\color{red}\implies \int_{a}^{b}f(x)dx\]
ach ... im used to i instead of k so theres some bleed over there
the picture is fine ...
k=0 to n, not a to b
does that seem familiar?
yes sort of let me try to write it out
1 to n is a right hand rule, 0 to n-1 is a left had rule ... im working from memory is all
either way... a=2, and b-a = 5 soo b=7 seems right
i would venture to say f(x) = x^3, but my brain is telling me to be ware of that assumption
how would you find k?
k is just the kth iteration of the partition
okay so what about n?
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