Question

PLEASE HELP!!!

What is the 4th term of (a-√2)^8

What is the 6th term of (2x-3y)^11

What is the 2nd term of (y-x)^4

PLEASE HELP!!!

Answer 1

Are you familiar with the Binomial Theorem, or at least Pascal's Triangle?

Answer 2

Yes to the binomial theorem but I have difficulty using it

Answer 3

Expand them

Answer 4

How though?

Answer 5

\[(a+b)^n=\sum_{k=0}^n\begin{pmatrix}n\\k\end{pmatrix}a^{n-k}b^k\]
The fourth term of \(\left(a-\sqrt2\right)^8\) is then
\[\begin{pmatrix}8\\3\end{pmatrix}a^{8-3}\left(\sqrt2\right)^3=\frac{8!}{3!5!}\times 2^{3/2}a^5=\cdots\]

Answer 6

Do the same for the rest. Keep track of the index: keep in mind that \(k=0\) gives you the first term, so \(k=1\) gives the second, and so on.

Answer 7

Ok so would the final answer be -112√(2) a^5 ... (I have multiple choice)