Respuesta :

frika

Answer:

Correct choice is C

Step-by-step explanation:

The i-th term of the binomial expansion [tex]\left(3x+7y\right)^{11}[/tex] is

[tex]T_i=C^{11}_{i-1}\cdot \left(3x\right)^{11+1-i}\cdot (7y)^{i-1}.[/tex]

If i=7, then

[tex]T_7=C^{11}_{7-1}\cdot \left(3x\right)^{11+1-7}\cdot (7y)^{7-1}=C_6^{11}\cdot (3x)^5\cdot (7y)^6=462\cdot (3x)^5\cdot (7y)^6.[/tex]

Note that

[tex]C_6^{11}=\dfrac{11!}{6!(11-6)!}=\dfrac{6!\cdot 7\cdot 8\cdot 9\cdot 10\cdot 11}{6!\cdot 1\cdot 2\cdot 3\cdot 4\cdot 5}=462.[/tex]

alinakincsem

Answer:

The correct answer option is C. [tex]462 (3x)^5 (7y)^6[/tex].

Step-by-step explanation:

We are given the following expression to be expanded and we are to find its seventh term:

[tex](3x+7y)^{11}[/tex]

The coefficient here is taken from Pascal's triangle (nCr on the calculator).

The expression expands in a way such that, the power of the first term decreases by one each time while it increases for the second term.

1st term: [tex]1 (3x)^11[/tex]

2nd term: [tex]11 (3x)^{10} (7y)^1[/tex]

3rd term: [tex]55 (3x)^9 (7y)^2[/tex] and so on.

Therefore, the 7th term will be [tex]462 (3x)^5 (7y)^6[/tex]