Question

The coefficient of x^3y^4 in (3x+2y)^7 is

Answer 1

Answer:

The coefficient is 15120.  

Step-by-step explanation:

Since, by the binomial expansion formula,

$(x+y)^n=\sum_{r=0}^n^nC_r x^{n-r} y^r$

Where, $^nC_r=\frac{n!}{r!(n-r)!}$

Thus, we can write,

$(3x+2y)^7 = \sum_{r=0}^n ^7C_r (3x)^{7-r} (2y)^r$

For finding the coefficient of $x^3y^4$,

r = 4,

So, the term that contains $x^3y^4$ = $^7C_4 (3x)^3 (2y)^4$

$=35 (27x^3) (16y^4)$

$=15120 x^3 y^4$

Hence, the coefficient of $x^3y^4$ is 15120.

Answer 2

Answer:

Coefficient of  $x^3y^4$ in $(3x+2y)^7$ is 15120

Step-by-step explanation:

We know that $(x+y)^{n}$) can be expanded in (n+1) terms by using binomial theorem and each term is given as

$n_C_{r}x^{n-r}y^{r}$

Here value of r is taken from n to 0

we have to determine the coefficient of $x^3y^4$ in $(3x+2y)^7$

in this problem we have given n=7

We have to determine the coefficient of $x^3y^4$

it means in the expansion we have to find the the 3rd power of x and therefore

r=n-3

here n=7

therefore, r=7-3=4

Hence the coefficient of $x^3y^4$  can be determine by using formula

$n_C_{r}x^{n-r}y^{r}$

here n=7, r=4

$7_C_{4}x^{7-4}y^{4}$

=$\frac{7\times 6\times 5\times 4}{1\times 2\times 3\times 4} (3x)^3(2y)^4$

=$15120x^3y^4$

Therefore the coefficient of  $x^3y^4$ in $(3x+2y)^7$ is 15120