Question 120295



{{{(3x+y)^5}}} Start with the given expression


To expand this, we're going to use binomial expansion. So let's look at Pascal's triangle:
<center>1&nbsp; &nbsp;</center>
<center>1&nbsp; &nbsp;1&nbsp; &nbsp;</center>
<center>1&nbsp; &nbsp;2&nbsp; &nbsp;1&nbsp; &nbsp;</center>
<center>1&nbsp; &nbsp;3&nbsp; &nbsp;3&nbsp; &nbsp;1&nbsp; &nbsp;</center>
<center>1&nbsp; &nbsp;4&nbsp; &nbsp;6&nbsp; &nbsp;4&nbsp; &nbsp;1&nbsp; &nbsp;</center>
<center>1&nbsp; &nbsp;5&nbsp; &nbsp;10&nbsp; &nbsp;10&nbsp; &nbsp;5&nbsp; &nbsp;1&nbsp; &nbsp;</center>




Looking at the row that starts with 1,5, etc, we can see that this row has the numbers:


1, 5, 10, 10, 5, and 1


These numbers will be the coefficients of our expansion. So to expand {{{(3x+y)^5}}}, simply follow this procedure:

Write the first coefficient. Multiply that coefficient with the first binomial term {{{3x}}} and then the second binomial term {{{y}}}. Repeat this until all of the coefficients have been written.


Once that has been done, add up every term like this:



{{{highlight(1)(3x)(y)+highlight(5)(3x)(y)+highlight(10)(3x)(y)+highlight(10)(3x)(y)+highlight(5)(3x)(y)+highlight(1)(3x)(y)}}} Notice how the coefficients are in front of each term.




However, we're not done yet.



{{{1(3x)^5(y)^0+(3x)(y)+10(3x)(y)+10(3x)(y)+5(3x)(y)+1(3x)(y)}}} Looking at the first term {{{1(3x)(y)}}}, raise  {{{3x}}} to the 5th power and raise {{{y}}} to the 0th power.


{{{1(3x)^5(y)^0+(3x)^4(y)^1+10(3x)(y)+10(3x)(y)+5(3x)(y)+1(3x)(y)}}} Looking at the  second term {{{5(3x)(y)}}} raise  {{{3x}}} to the 4th power and raise {{{y}}} to the 1st power.


Continue this until you reach the final term.



Notice how the exponents of {{{3x}}} are stepping down and the exponents of {{{y}}}  are stepping up.



So the fully expanded expression should now look like this:



{{{1(3x)^5(y)^0+5(3x)^4(y)^1+10(3x)^3(y)^2+10(3x)^2(y)^3+5(3x)^1(y)^4+1(3x)^0(y)^5}}}



{{{1(243x^5)(y^0)+5(81x^4)(y^1)+10(27x^3)(y^2)+10(9x^2)(y^3)+5(3x^1)(y^4)+1(x^0)(y^5)}}} Distribute the exponents



{{{1(243x^5)+5(81x^4y)+10(27x^3y^2)+10(9x^2y^3)+5(3xy^4)+1(y^5)}}} Multiply



{{{243x^5+405x^4y+270x^3y^2+90x^2y^3+15xy^4+y^5}}} Multiply the terms with their coefficients



So {{{(3x+y)^5}}} expands and simplifies to {{{243x^5+405x^4y+270x^3y^2+90x^2y^3+15xy^4+y^5}}}.



In other words, {{{(3x+y)^5=243x^5+405x^4y+270x^3y^2+90x^2y^3+15xy^4+y^5}}}