Question

Expand (2 + 3x)^{−2} in ascending powers of x, up to and including the term in x^2, simplifying the
coefficients.

Answer 1

Binomial expansion? Have you found the General Term?

Answer 2

\[(1+x)^n = 1+nx + {n(n-1)x^2 \over 2!}...\]

Answer 3

How to get it into 1 + x and n is what? Can you show me how to solve this completely?

Answer 4

You bring 2 outside the brackets.

Answer 5

so it becomes 2(1 + 3/2x)^-2

Answer 6

First..isn't the expansion of (1+x)^n
\[(1+x)^n = \left(\begin{matrix}n \\ 0\end{matrix}\right)+\left(\begin{matrix}n \\ 1\end{matrix}\right)x+\left(\begin{matrix}n \\ 2\end{matrix}\right)x^2+...+\left(\begin{matrix}n \\ r\end{matrix}\right)x^r+..+\left(\begin{matrix}n \\ n\end{matrix}\right)x^n\]

Answer 7

Not if n = <1

Answer 8

\[2(1 + 3/2x)^{-2} ~~~~~~You~have~ t o~use~this\]

Answer 9

and the formula I put up

Answer 10

The first term will be 1/4, but not sure how to get that..

Answer 11

\[(2+3x)^{-2}\]
\[(2)^{-2}(1+\frac{3}{2}x)^{-2}\]
\[n=-2 , x_1=\frac{3}{2}x\]
Expand
\[\frac{1}{4}[1+(-2)(\frac{3}{2}x)+\frac{-2(-2-1)}{2}(\frac{3}{2}x)^2]\]
\[\frac{1}{4}(1-3x+\frac{27}{4}x^2)\]

Answer 12

Thanks so much! This clears it up. I forgot about the ^-2 for the 2