How do you use the Binomial theorem to expand (3x+y^2)^7(3x+y2)7?

1 Answer
Aug 25, 2017

(3x+y^2)^7=2187x^7+5103x^6y^2+5103x^5y^4+2835x^4y^6+945x^3y^8+189x^2y^10+21xy^12+y^14(3x+y2)7=2187x7+5103x6y2+5103x5y4+2835x4y6+945x3y8+189x2y10+21xy12+y14

Explanation:

According to binomial theorem expansion of (a+b)^n(a+b)n

= C_0^na^n+C_1^na^(n-1)b+C_2^na^(n-2)b^2+....+C_r^na^(n-r)b^r+....+C_(n-1)^nab^(n-1)+C_n^nb^n

where C_r^n=(n!)/(r!(n-r)!)=(n(n-1)(n-2)....(n-r+1))/(1*2*3*.......*r), where C_0^n=1

Also observe that C_(n-r)^n=C_r^n

Hence (3x+y^2)^7

= C_0^7(3x)^7+C_1^7(3x)^6(y^2)+C_2^7(3x)^5(y^2)^2+C_3^7(3x)^4(y^2)^3+C_4^7(3x)^3(y^2)^4+C_5^7(3x)^2(y^2)^5+C_6^7(3x)(y^2)^6+C_7^7(y^2)^7

= 2187x^7+5103x^6y^2+5103x^5y^4+2835x^4y^6+945x^3y^8+189x^2y^10+21xy^12+y^14

Observe that C_0^7=1, C_1^7=7, C_2^7=21, C_3^7=35, C_4^7=35, C_5^7=21, C_6^7=7 and C_7^7=1 and that these vaues can also beobtained using Pascal's triangle.