Question

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

Answer 1

`(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`

Explanation:

According to binomial theorem expansion of `(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.