Question

Determine terms 1-5 of the binomial expression (x+y)^4 . Once finished , write your answer in standard form using the answers , enter expanded form of the binomial expression (x+y)^4

Answer 1

The first term in the binomial, starts off with a highest exponent possible, in this case 4, and then it drops down by 1 every expanded term, till it gets to 0.

the second term in the binomial, starts at 0 and graduates its exponent by 1 on every term.

so, that's peanuts, the part the changes most is the coefficients.

one way to get the coefficients is sequentially, namely as you go

$\bf ~~~~~~~~\textit{binomial theorem expansion} \\\\ (x+y)^4\implies \begin{array}{cccllll} term&coefficient&value\\ ---&-----&-----\\ 1&+1&(x)^4(+y)^0\\ 2&+4&(x)^3(+y)^1\\ 3&+6&(x)^2(+y)^2\\ 4&+4&(x)^1(+y)^3\\ 5&+1&(x)^0(+y)^4 \end{array} \\\\\\ 1x^4y^0+4x^3y^1+6x^2y^2+4x^1y^3+1x^0y^4 \\\\\\ x^4+4x^3y+6x^2y^2+4xy^3+y^4$

so, the first coefficient is always 1, how do we get the next one?

well, "I multiply the current coefficient by the exponent of the first term, and divide that product by the exponent of the second on the following term", well, that's a mouthful, so let's see how's done.

to get the 2nd one of "4", we get it by (1*4)/1

to get the 3rd one of "6", is (4*3)/2

to get the 4th one of "4", is (6*2)/3

to get the 5th one of "1", is (4*1)/4

now, there's a more direct method to get any coefficient and terms

$\bf \textit{the coefficient and values of an expanded term} \\\\ (a-b)^n \qquad \qquad k=0,1,2,3,..,n-1\\\\ \stackrel{coefficient}{\left(\frac{n!}{k!(n-k)!}\right)} \qquad \stackrel{\stackrel{first~term}{factor}}{\left( a^{n-k} \right)} \qquad \stackrel{\stackrel{second~term}{factor}}{\left( b^k \right)}$

where "k" is a value of the term starting at 0.

so, to get say with this method the 3rd term

$\bf (x+y)^4 \qquad \qquad \begin{array}{llll} expansion\\ for\\ 3^{th}~term \end{array} \quad \begin{cases} k=\stackrel{3rd~term}{2}\\ n=4 \end{cases}\\\\ \stackrel{coefficient}{\left(\frac{4!}{2!(4-2)!}\right)} \qquad \stackrel{\stackrel{first~term}{factor}}{\left( x^{4-2} \right)} \qquad \stackrel{\stackrel{second~term}{factor}}{\left( y^2 \right)} \\\\\\ \cfrac{4!}{2!2!}~x^2y^2\implies \cfrac{3\cdot 4}{2!}x^2y^2\implies \cfrac{12}{2}x^2y^2\implies 6x^2y^2$