Answer:
The coefficient of x³y⁴ in the expansion of ( x+2y)⁷ is 560.
Explanation:
The given expression is
[tex](x+2y)^7[/tex]
According to binomial expansion,
[tex](a+b)^n=^nC_0a^nb^0+^nC_1a^{n-1}b^1+...+^nC_{n-1}a^1b^{n-1}+^nC_0a^0b^n[/tex]
The r+1th term of the expansion is
[tex]^nC_rx^{n-r}(2y)^r=^nC_r(2^r)x^{n-r}(y)^r[/tex] ... (1)
In the term x³y⁴ the power of x is 3 and the power of y is 4. It means the value of r is 4 and the value n-r is 3.
[tex]n-r=3[/tex]
[tex]n-4=3\Rightarrow n=7[/tex]
Put n=7 and r=4 in equation (1)
[tex]^7C_4(2^4)x^{7-4}(y)^4[/tex]
[tex]\frac{7!}{4!(7-4)!}(16)x^3y^4[/tex]
[tex]\frac{7\times 6\times 5\times 4!}{4!(3)!}(16)x^3y^4[/tex]
[tex]560x^3y^4[/tex]
Therefore the coefficient of x³y⁴ in the expansion of ( x+2y)⁷ is 560.
