Question

Solve (4x+y^5)^7 using binomial theorem​

Answer 1

Step-by-step explanation:

$\text{The binomial theorem:}\\\\(a+b)^n = \sum\limits_{k=0}^n {n \choose k}a^{n-k}b^k\\\\{n \choose k}=\dfrac{n!}{k!(n-k)!}$

$\text{We have}\ (4x+y^5)^7\Rightarrow a=4x,\ b=y^5,\ n=7.\\\\\text{Substitute:}\\\\(4x+y^5)^7={7\choose 0}(4x)^{7-0}(y^5)^0+{7\choose 1}(4x)^{7-1}(y^5)^1+{7\choose 2}(4x)^{7-2}(y^5)^2\\\\+{7\choose 3}(4x)^{7-3}(y^5)^3+{7\choose 4}(4x)^{7-4}(y^5)^4+{7\choose 5}(4x)^{7-5}(y^5)^5\\\\+{7\choose 6}(4x)^{7-6}(y^5)^6+{7\choose 7}(4x)^{7-7}(y^5)^7\\\\=(1)(4x)^7(y^5)^0+(7)(4x)^6(y^5)^1+(21)(4x)^5(y^5)^2+(35)(4x)^4(y^5)^3\\\\+(35)(4x)^3(y^5)^4+(21)(4x)^2(y^5)^5+(7)(4x)^1(y^5)^6+(1)(4x)^0(y^5)^7$

$=16384x^7+28672x^6+21504x^5y^{10}+8960x^4y^{15}\\\\+2240x^3y^{20}+336x^2y^{25}+28xy^{30}+y^{35}$

$\text{Used}\\\\(a^n)^m=a^{nm}\\\\a^0=1$