# Using Pascal's Triangle for Binomial Expansion.

I'm trying to answer a question using Pascal's triangle to expand binomial functions, and I know how to do it for cases such as (x+1) which is quite simple, but I'm having troubles understanding and looking for a pattern when the question changes to (x+2)^n

I understand that (x+1)^3 would be x^3 + 3x^2 + 3x + 1 ; Using pascal's triangle

But I do not understand how to expand (x+2)^ 3 using pascal's triangle; I'm not sure what to do with the constant value 2 or any other value aside from 1 in this case.

• $(x+y)^3=x^3+3x^2y+3xy^2+y^3$. Put $y=2$. – André Nicolas Sep 10 '14 at 4:47

$$(x+1)^3=1(x^3)(1^0)+3(x^2)(1^1)+3(x^1)(1^2)+1(x^0)(1^3)$$ If you can discern the pattern, it's a logical step toward being able to expand $(x+2)^3$.
To proceed from $x\to x+1$ to $x\to x+2$and to $x \to x+a$ yand to powers of it you can consider powers of the pascal Matrix. You might lookout for my articles on the Pascal matrix.
Here is a matrix-multiplication-scheme involving the Pascalmatrix $P$:
It says, that with a column-vector with consecutive powers of an indeterminate $s$ the so-to-say "Pascal-transformation" is a vector with consecutive powers of $s+1$. Or more formally said:
Assume a type of columnvector $V(x) = [1,x,x^2,x^3,...,x^n]$ then $$P \cdot V(x) = V(x+1)$$ Not very surprising $$P \cdot V(x+1) = V(x+2)$$ or $$P \cdot (P \cdot V(x)) = V(x+2)$$ or $$P^2 \cdot V(x) = V(x+2)$$ and in general $$P^h \cdot V(x) = V(x+h)$$ It is then interesting, that the powers of $P$ can be written by three rather easy methods.
After this is now so easy... might you get an idea how to define negative, or even fractional, $h$?