# Using pascal's triangle how do you solve the equation (2x+3y)^6. I am not sure how this correlates when there are coefficients.Is this correct?...

Using pascal's triangle how do you solve the equation (2x+3y)^6. I am not sure how this correlates when there are coefficients.

Is this correct?

64x^6+576x^5y+2160x^4y^2+4320x^3y^3+4860x^2y^4+2916xy^5+729y^6

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Expand the expression `(2x+3y)^6` using Pascal's triangle:

(1) The coefficients for each term can be found using the appropriate line of Pascal's triangle:

1 `(a+b)^0=1`

1 1 `(a+b)^1=1a+1b`

1 2 1 `(a+b)^2=1a^2+2ab+1b^2`

etc... For sixth power the coefficients are 1 6 15 20 15 6 1

(2) The expansion is:

`(2x+3y)^6=1(2x)^6(3y)^0+6(2x)^5(3y)^1+15(2x)^4(3y)^2+20(2x)^3(3y)^3` `+15(2x)^2(3y)^4+6(2x)^1(3y)^5+1(2x)^0(3y)^6`

`=64x^6+576x^5y+2160x^4y^2+4320x^3y^3+4860x^2y^4'`

`+2916xy^5+729y^6`