Question

find the binomial expansion of (x-1/x^3)^10

Answer 1

oh well lets start with the first rule about exponents

Answer 2

x^10-10x^6+45x^2-120/x^2+210/x^6-252/x^10 +210/x^14 -120/x^18+45/x^22-10/x^26+1/x^30

Answer 3

okay i seen the answer too in wolfram alpha i need to understand not the answer lol

Answer 4

Binomial theorem:
\[\large(a+b)^n=\sum_{k=0}^na^kb^{n-k}\]
Set \(a=x\) and \(b=-\dfrac{1}{x^3}\).

Answer 5

Oh and \(n=10\).

Answer 6

okay so if its x-1/x^3

Answer 7

i should jus rearrange it to be -1/x^3 + x correct

Answer 8

That shouldn't matter.

Answer 9

\((a+b)^n=(b+a)^n\)

Answer 10

hmmm ill write that down ..

Answer 11

how will i get the coefficents though

Answer 12

That's where the binomial coefficient comes in. The first term, for \(k=0\), would be
\[\binom{10}0x^0\left(-\frac{1}{x^3}\right)^{10-0}=\frac{10!}{0!(10-0)!}\times1\times\left(-\frac{1}{x^3}\right)^{10}=\frac{1}{x^{30}}\]

Answer 13

okay thx so the coefficient is just (n-1
k )
right

Answer 14

oh no its

Answer 15

Not sure where you're getting \(n-1\) from. Here's the def. of the binomial coefficient:
\[\binom nk=\frac{n!}{k!(n-k)!}\]

Answer 16

yea i saw it was wrong was thinking something else

Answer 17

thx

Answer 18

yw