Calculating the Coefficient of x^34 in (1+x^2+x^7+x^16)^1000

[FlyingMonkey]

(1+x^2+x^7+x^16)^n
What is the coeffient of x^34 when n=1000?

thanku

 

[gazzo]

what have you done so far?

 

[FlyingMonkey]

what have you done so far?

what ? so far ?
i stay at one place only. no farther, very near!
means i don no how to solve problem.

 

[HallsofIvy]

what ? so far ?
i stay at one place only. no farther, very near!
means i don no how to solve problem.

Do you know anything about the "binomial theorem" or "multinomial theorem". If not, why in the world are you doing a problem like this?

 

[Parth Dave]

Speaking of the multinomial theorem, does anyone know where i can get sample problems? I have a test next tuesday that is on the binomial and multinomial theorem, unfortunately my textbook has none.

 

[TenaliRaman]

Speaking of the multinomial theorem, does anyone know where i can get sample problems? I have a test next tuesday that is on the binomial and multinomial theorem, unfortunately my textbook has none.

You may like to get your hands on Hall & Knight, if its possible for u to get it through some library maybe.

 

[robert Ihnot]

(1+x^2+x^7+x^16)^n
What is the coeffient of x^34 when n=1000?

thanku

Well, to present a "hands on" start on this, you have to look at all the ways you can come up with X^34, and the exponents on these algebraic terms total 1000. (We need a lot of 1s.) One way is (X^16)^2(X^2)^1(1^997). (Hopefully you understand what I mean.) Coefficient for this is then (1000!)/[2!1!997!].

Then we proceed to look at (X^16)^1(X^7)^2(X^2)^2(1^997), this gives us the coefficient: (1000!)/[1!2!997!].

So proceeding in this method to look at all cases, we then add up all the coefficients, and presumedly, that is the correct answer.

A simpler question would be to find the coefficient for X^3 on (1+X+X^2)^4 .

There is only two ways: (X)^1(X^2)^1(1^2), and (x)^3(1^1). So the coefficient is 4!/(2!1!1!) and 4!/(3!1!) = 12 + 4 = 16.

Check out: http://encyclopedia.thefreedictionary.com/Trinomial theorem