What is the sum of all of the coefficients $(4x-3)^{10}$? The answer is $1$ but I need a faster way of doing this. Right now I just expanded it all using Pascal's triangle and $n$ choose $k$. I took this down to $(4x-3)^{4}$ and got $1$ as well. I then did $(4x-2)^{4}$ and got $16$. So is there a formula for finding the sum of coefficients for $(ax-by)^{n}$ just as $(a-b)^n$?
What about when it is $(ax+by)^{n}$?
On
There is, I haven't really mastered the art of LaTeX, so I won't be able to write the exact solution, but, the idea is as follows:
Use Binomial Theorem: http://en.wikipedia.org/wiki/Binomial_theorem
Supplement it with a "Sum from k to n" for the coefficient. Simplify the equation.
If you do that, you're going to notice that this solves both the original question, and the question of $(ax+by)^c$
The sum of all the coefficients should equal to $p(1) = (4\cdot 1 - 3)^{10} = 1$ with $p(x) = (4x - 3)^{10}$