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How do you solve (2x+y)^5 using Pascal’s Triangle?

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check the picture below.

then recall that the first term start with a highest exponent and gradually wanes, and the second term starts at 0, and gradually goes up. Then using those coefficients, expand away.



\bf ( 2x+y )^5 \\\\\\ 1( 2x )^5(y)^0~+~5( 2x )^4( y )^1~+~10( 2x )^3( y )^2~+~10( 2x )^2( y )^3\\\\+~5( 2x )^1( y )^4~+~1( 2x )^0( y )^5 \\\\\\ 2^5x^5+5(2^4x^4)(y)+10(2^3x^3)(y^2)+10(2^2x^2)(y^3)\\\\ +5(2^1x^1)(y^4)+1(1)(y^5) \\\\\\ 32x^5+80x^4y+80x^3y^2+40x^2y^3+10xy^4+y^5
How do you solve (2x+y)^5 using Pascal’s Triangle?-example-1
User Wolfgang Fahl
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