Solving the Coefficient of x^33 in Binominal Theorem Expansion

[morbello]

ive been asked to work out the coefficent of x^33 in the expansion of

(1/4 2x^3)^17

i know about (a+b) and ^C_ but I am unsure how to get k which is c_k

i worked off (1/4)^17-k (2x^3)^k

which give me 17*1/4 =4 1/4 and that is were i went wrong could you tell me a way to work out k

Homework Equations

The Attempt at a Solution

 

[danago]

What is the only value of k that will produce the x^33 term?

 

[morbello]

so your saying some thing as simple as -16 would make the sum work.

 

[danago]

Im not sure how you got -16.

The binomial expansion formula says that:

$$(0.25+2x^3)^{17} = \sum \!^{17} C _k (0.25)^{17-k} (2x^3)^k$$

As k ranges from 0 to 17. For which of those values of k will the x^33 term appear?

 

[morbello]

that is what I am not sure about i don't know an easy way to find k

 

[danago]

Ok, well consider a few different values for k.

If k=1, then the (x^3)¹=x³ term will appear. If k=2, then the (x^3)²=x⁶ term appears. Can you see now what value of k will make the x^33 term appears?

 

[morbello]

ok i see what you mean i did see a pice in the book about it but i was not sure what it ment.thank you i was thinking that both a+b worked together and did my maths around that i see i could look at it as b on its own.