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Which expression correctly shows how to use the binomial theorem to determine the 4th term in the expansion of (2x^2y^3+y)^7 ?

Must be correct the highlighted is my answer I need to know if it's right or wrong.

Which expression correctly shows how to use the binomial theorem to determine the 4th term in the expansion of 2x2y3y7 Must be correct the highlighted is my ans class=

Respuesta :

Answer:  Choice D) 560x^8y^15

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Explanation:

We dont involve sigma notation since we want just one term and we aren't adding up a bunch of things. It's really curious why your teacher is using sigma notation as it is incorrect. Everything else is fine.

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n = 7 is the outer most exponent of the binomial.

We have the terms for k = 0, k = 1, ... which means the 4th term is when k = 3.

n C k = (n!)/(k!*(n-k)!)

7 C 3 = (7!)/(3!*(7-3)!)

7 C 3 = (7!)/(3!4!)

So that explains the coefficient of (7!)/(3!4!). This is known as a binomial coefficient. Some books call the n C k function the "choose" function. Though other books call it a combination function. They're the same thing. Let's keep going to fully compute the binomial coefficient.

7 C 3 = (7!)/(3!4!)

7 C 3 = (7*6*5*4*3*2*1)/(3*2*1*4*3*2*1)

7 C 3 = (5040)/(6*24)

7 C 3 = (5040)/(144)

7 C 3 = 35

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a = 2x^2y^3

b = y

The kth term is in the form

(n C k)*(a)^(n-k)*(b)^k

Plug in the given values and simplify

(7 C 3)*(a)^(7-3)*(b)^3

35*(a)^4*(b)^3

35*(2x^2y^3)^4*(y)^3

35*16*(x^8y^12)*(y)^3

560*x^8y^12*y^3

560*x^8y^(12+3)

560x^8y^15

sqdancefan

Answer:

  you need to rethink your selection

Step-by-step explanation:

None of the answer choices is correct. The fourth term does not need or have a summation symbol. (The effect of that symbol as written here is to multiply the term by 8.)

In (a+b)^7, the powers of "a" count down from 7 to 0. The fourth term has a^4b^3, so matches the left-side exponents of the first and last choices. Unfortunately, the right-side exponents of choice A don't match those on the left, so the only appropriate choice here is choice D.

As you know, the coefficient will be ...

  C(7, 3)(2^4) = 35×16 = 560

The fourth term of the expansion is ...

  560x^8·y^15

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When expressed with a summation symbol, the entire expansion (not just a single term) is ...

  [tex]\sum\limits_{k=0}^{7}{\left(\dfrac{7!}{k!(7-k)!}(2x^2y^3)^{7-k}y^k\right)}[/tex]

The fourth term has k=3.

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