Respuesta :
Answer:
The coefficient is 15120.
Step-by-step explanation:
Since, by the binomial expansion formula,
[tex](x+y)^n=\sum_{r=0}^n^nC_r x^{n-r} y^r[/tex]
Where, [tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]
Thus, we can write,
[tex](3x+2y)^7 = \sum_{r=0}^n ^7C_r (3x)^{7-r} (2y)^r[/tex]
For finding the coefficient of [tex]x^3y^4[/tex],
r = 4,
So, the term that contains [tex]x^3y^4[/tex] = [tex]^7C_4 (3x)^3 (2y)^4[/tex]
[tex]=35 (27x^3) (16y^4)[/tex]
[tex]=15120 x^3 y^4[/tex]
Hence, the coefficient of [tex]x^3y^4[/tex] is 15120.
Answer:[tex][/tex]
Coefficient of [tex]x^3y^4[/tex] in [tex](3x+2y)^7[/tex] is 15120
Step-by-step explanation:
We know that [tex](x+y)^{n}[/tex]) can be expanded in (n+1) terms by using binomial theorem and each term is given as
[tex]n_C_{r}x^{n-r}y^{r}[/tex]
Here value of r is taken from n to 0
we have to determine the coefficient of [tex]x^3y^4[/tex] in [tex](3x+2y)^7[/tex]
in this problem we have given n=7
We have to determine the coefficient of [tex]x^3y^4[/tex]
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 [tex]x^3y^4[/tex] can be determine by using formula
[tex]n_C_{r}x^{n-r}y^{r}[/tex]
here n=7, r=4
[tex]7_C_{4}x^{7-4}y^{4}[/tex]
=[tex]\frac{7\times 6\times 5\times 4}{1\times 2\times 3\times 4} (3x)^3(2y)^4[/tex]
=[tex]15120x^3y^4[/tex]
Therefore the coefficient of [tex]x^3y^4[/tex] in [tex](3x+2y)^7[/tex] is 15120
