Question

The largest term in this sequence=?

Answer 1

\[\Huge a_n=\frac{n^2}{n^3+200}\]

Answer 2

@Callisto @oldrin.bataku

Answer 3

I think for An to be maximum denominator has to minimum?maybe?

Answer 4

@DLS denominator has to be minimum

Answer 5

The maximum should happen near the maximum of \(f(x)=\frac{x^2}{x^3+200}\).

Answer 6

\[\LARGE Let~f(x)=n^3+200\]
\[\LARGE f'(x)=3n^2\]
how to proceed after this?

Answer 7

... since your function is continuous and all

Answer 8

$$f(x)=\frac{x^2}{x^3+200}\\f'(x)=\frac{2x(x^3+200)-x^2(3x^2)}{\cdots}\\0=2x(x^3+200)-x^2(3x^2)\\0=2x^4+400x-3x^4\\x^4-400x=0\\x(x^3-400)=0\\x^3-400=0\\x^3=400\\x=\sqrt[3]{400}$$For \(n\), consider the maximum of \(f(\lfloor\sqrt[3]{400}\rfloor),f(\lceil\sqrt[3]{400}\rceil)\)

Answer 9

I have options..
529/49
8/89
49/543
None of these..

Answer 10

It's one of those listed

Answer 11

i have to substitute n=cube root of 400?

Answer 12

No because \(n\) must be an integer. Do you not understand floor/ceiling notation?

Answer 13

oh yeah didn't see that bracket,sorry.
Why should it be an integer?

Answer 14

Observe: \(\sqrt[3]{400}=7.3\dots\) hence \(\lfloor\sqrt[3]{400}\rfloor=7\) while \(\lceil\sqrt[3]{400}\rceil=8\). Our maximum term is either \(a_7\) or \(a_8\)

Answer 15

@DLS a sequence \(a_n\) is defined as a function whose domain is the nonnegative integers usually...

Answer 16

so we will go with a8?

Answer 17

Did you compute both?

Answer 18

because no \(a_8\) is actually the smaller of the two :-p

Answer 19

a7=49/543
a8=64/712=32/356=16/178=8/89

Answer 20

so a7 is greater?

Answer 21

why did we differentiate whole thing and not only the denominator?

Answer 22

Indeed, \(a_7\) is the greatest term in the sequence.

Answer 23

hmm,okay!thanks!

Answer 24

The idea is that, since the function corresponding to our sequence is continuous, we expect them to have maximum values in the same vicinity: