Question

Expand (4x-3y)^4 using pascal's triangle

Answer 1

Well with the fourth row of pascals triangle, we have:
\[1\phantom{next}4\phantom{next}6\phantom{next}4\phantom{next}1\]

So we obtain the formula:
\[(a+b)^n=\sum_{m=0}^n{P[n+1]_{m+1}(a)^{n-m}(b)^{m}}\]
Where \(P[n]_m\) is the \(m\)th term of the \(n\)th column of the pascals trangle.
And we can use this for let's say something in the form of a quadratic like so:
\[\eqalign{
(a+b)^2&=\sum^2_{m=0}P[3]_{m+1}a^{2-m}b^m \\
&=P[3]_1a^2b^0+P[3]_2a^1b^1+P[3]_3a^0b^2 \\
&=a^2+2ab+b^2 \\
}\]
In our case:

Answer 2

We have:
\[\eqalign{
(a+b)^4&=\sum^4_{m=0}P[5]_{m+1}a^{4-m}b^m \\
&=P[5]_1a^4b^0+P[5]_2a^3b^1+P[5]_3a^2b^2+P[5]_4a^1b^3+P[5]_5a^0b^4 \\
&=a^4+4a^3b+6a^2b^2+4ab^3+b^4
}\]

Answer 3

Let us make the substitution:
\[\eqalign{
&a=4x \\
&b=-3y
}\]
And we obtain:
\[\eqalign{
&a^4+4a^3b+6a^2b^2+4ab^3+b^4 \\
&=(4x)^4+4(4x)^3(-3y)+6(4x)^2(-3y)^2+4(4x)(-3y)^3+(-3y)^4 \\
&=256x^4-768x^3y+864x^2y^2-432xy^3+81y^4 \\
}\]

Answer 4

thank you i did not know what i should have done with the exponents

Answer 5

Haha well I hope that that's not only a solution for ya now but also if you ever have to expand another one. It WAS a pretty general answer then I made it specific!