Question

$(2x + 4y)^4$
Expanded using Pascal's triangle or binomial theorem ​

Answer 1

You build Pascal's triangle as follows: every row starts and ends with 1, and every term is the sum of the two immediately above. So, we have

$\begin{array}{c}n=0\\n=1\\n=2\\n=3\\n=4\end{array}\qquad\begin{array}{ccccccccc}&&&&1&&&&\\&&&1&&1&&&\\&&1&&2&&1&&\\&1&&3&&3&&1&\\1&&4&&6&&4&&1\end{array}$

We can use this triangle to expand powers as follows: the binomial $(a+b)^n$ will feature all the powers $a^xb^y$, where $x+y=n$, with some coefficients: in this case, we have

$(a+b)^4 = c_4a^4b^0 + c_3a^3b^1+c_2a^2b^2+c_1a^1b^3+c_0a^0b^4$

and the coefficients $c_i$ are exactly the required row of the Pascal's triangle. Removing also unnecessary exponents, we have

$(a+b)^4 = a^4 + 4a^3b+6a^2b^2+4ab^3+b^4$

Finally, substitute a=2x and b=4y to get the answer:

$(2x+4y)^4 = 16 x^4 + 128 x^3 y + 384 x^2 y^2 + 512 x y^3 + 256 y^4$