Question

Find the coefficient of x3 y4 in the expansion of ( x+2y )^7.

Answer 1

Answer:

The coefficient of x³y⁴ in the expansion of ( x+2y)⁷ is 560.

Explanation:

The given expression is

$(x+2y)^7$

According to binomial expansion,

$(a+b)^n=^nC_0a^nb^0+^nC_1a^{n-1}b^1+...+^nC_{n-1}a^1b^{n-1}+^nC_0a^0b^n$

The r+1th term of the expansion is

$^nC_rx^{n-r}(2y)^r=^nC_r(2^r)x^{n-r}(y)^r$       ... (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.

$n-r=3$

$n-4=3\Rightarrow n=7$

Put n=7 and r=4 in equation (1)

$^7C_4(2^4)x^{7-4}(y)^4$

$\frac{7!}{4!(7-4)!}(16)x^3y^4$

$\frac{7\times 6\times 5\times 4!}{4!(3)!}(16)x^3y^4$

$560x^3y^4$

Therefore the coefficient of x³y⁴ in the expansion of ( x+2y)⁷ is 560.