Question 1039773: this is about a binomial theorem...
SOLVE: (3x+2y)^10
solve it in binomial theorem.Thanks.I really need your help.I ALSO,needs solution or step by step for it.Thanks! :) I hopeyou could help me. :)))
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The 10 terms have coefficients of 10C1, 10C2...10C10
each term has the coefficient multiplied by (3x) raised to the 10,9,8,7,... and (2y) raised to the 0,1,2,3,4,...
The total of the exponents will equal 10. Remember to raise the 3 and the 2 as well.
(3x+2y)^10=
10C0(3x)^10*2y^0=3^10*x^10+10*3^9x^9(2y)+45(10C3)*3^8x^8*4y^2(2y)^2+120*3^7x^7(8y^3)+
210*(3^6)(x^6)(16y^4)+252(3^5)x^5(32y^5)+210(81x^4)(64y^6)+120(27x^3)(128y^7)+45(9x^2)(256y^8)+
10(3x)(512y^9)+1024y^10
10Cn((x term)^n*(y term)^(10-n)
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