Question

Describe the positive integer values of n for which (x^2+\frac{1}{x})^n has a nonzero constant term. In terms of n, what is that constant term? (You can leave your answer in the form of a combination.)

Answer 1

can you type the expression in equation editor or write it out on the draw please?

Answer 2

wait

Answer 3

\[\Big(x^2+\frac{1}{x}\Big)^n\]

Answer 4

do a few... staring with n = 1 => no constant term

you should start to see a pattern.

binomial theorem says
\[\left( a+b \right)^{n}= \left(\begin{matrix}n \\ 0\end{matrix}\right)a ^{n}b ^{0}+\left(\begin{matrix}n \\ 1\end{matrix}\right)a ^{n-1}b ^{1}+\left(\begin{matrix}n \\ 2\end{matrix}\right)a ^{n-2}b ^{2}+...+\left(\begin{matrix}n \\ r\end{matrix}\right)a ^{n-r}b ^{r}+...+\left(\begin{matrix}n \\ n\end{matrix}\right)a ^{n-n}b ^{n}\]

in your case, you'll need this to have a constant term
\[\left( x ^{2} \right)^{k}\left( \frac{ 1 }{x } \right)^{2k}=1\]
so that the constant term will be
\[\left(\begin{matrix}2k+k \\ 2k\end{matrix}\right)\]
so n must be...

Answer 5

does that make any sense? you gonna reply?

Answer 6

uhh...

Answer 7

uhh?

Answer 8

still reading through it

Answer 9

okay... if you have any questions or need any clarification... please let me know

Answer 10

does that mean that it must be an even number?

Answer 11

what must be an even number?

Answer 12

the constant term

Answer 13

no... 3 chose 2 isn't even

Answer 14

2k+k = n so n must be what kind of number?

Answer 15

btw, k = 1, 2, 3, ...

Answer 16

an odd number

Answer 17

nope... what's 2k + k?

Answer 18

3k

Answer 19

oh. so its a multiple of 3!!!

Answer 20

I'm such an idiot

Answer 21

it's okay

Answer 22

why did I not see that? you're really smart pgpilot

Answer 23

i wish, but thanks
give yourself some cred as well and don't get down on you!

Answer 24

ok thank you so much!!

Answer 25

you are more than welcome!

Answer 26

keep the good questions like that one coming... my brain needs the challenge!