1. ## Binomial expansion help

i have two questions i need help with.
for the first one should i use pascals triangle?

1. Expand, fully, each of the following
i) (x + 2y)^4
2. Show that, if x is small enough for x^2 and higher powers of x to be neglected, the function (x – 2)(1 + 3x)^8 has a linear approximation of - 2 – 47x.

2. Originally Posted by perryman
i have two questions i need help with.
for the first one should i use pascals triangle?

1. Expand, fully, each of the following
i) (x + 2y)^4
2. Show that, if x is small enough for x^2 and higher powers of x to be neglected, the function (x – 2)(1 + 3x)^8 has a linear approximation of - 2 – 47x.
If you like, I find it easier to use factorials

$\displaystyle (a+b)^n = {n \choose 0}a^nb^0 + {n \choose 1}a^{n-1}b + {n \choose 2}a^{n-2}b^2 + {n \choose 3}a^{n-3}b^3 + ... + {n \choose n}a^0b^n$

Of course $\displaystyle {n \choose 0} = {n \choose n} = b^0 = a^0 = 1$

3. generally...
$\displaystyle (x+y)^n=\sum_{k=0}^{n}\begin{pmatrix} n\\ k \end{pmatrix}x^{n-k}y^{k}$

4. for the first question i got the answer $\displaystyle x^4 + (8x^3 * y) + (24x^2 *y^2) +32xy^3 + 16y^4$ can anyone confirm this

5. Hello perryman
Originally Posted by perryman
i have two questions i need help with.
for the first one should i use pascals triangle?

1. Expand, fully, each of the following
i) (x + 2y)^4
2. Show that, if x is small enough for x^2 and higher powers of x to be neglected, the function (x – 2)(1 + 3x)^8 has a linear approximation of - 2 – 47x.
e^(i*pi) and Raoh have given you the formula you need to expand #1. Do you understand how to use it?

To give you an example, and to help you with #2, I'll expand $\displaystyle (1+3x)^8$, ignoring powers of $\displaystyle x$ above $\displaystyle x^2$, using the formula the others have given you:
$\displaystyle (1+3x)^8 = \binom801^8(3x)^0 + \binom811^7(3x)^1 +\binom821^6(3x)^2 + ...$
Looks scary, doesn't it? But remember that $\displaystyle 1^n = 1$ for all values of $\displaystyle n,\; (3x)^0 = 1$ and
$\displaystyle \binom80 = 1$

$\displaystyle \binom81=8$

$\displaystyle \binom82 = \frac{8.7}{2!}$

So, it's simply:
$\displaystyle 1 +8(3x) + \frac{8.7}{2!}(3x)^2 + ...$
$\displaystyle =1+24x+...$ (and I've left the $\displaystyle x^2$ term out now, because we don't need it anyway - I just put it there as an example)
So if we ignore $\displaystyle x^2$ and higher powers of $\displaystyle x$:
$\displaystyle (x-2)(1+3x)^8 = (x-2)(1+24 x + ...)$
Can you complete it now?

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### show that if x is small enough for it cube and higher powers to be neglected

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