Calculating Coefficient of u^16v^2 in (u^2 + v)^10 Expansion

In summary: But you can't just use the combination of the two terms.In summary, the coefficient of u^16v^2 in the binomial expansion of (u^2 + v)^10 is 45. This is found by recognizing that n = 10 and b is raised to the second power, and using the combination formula (n choose 2). Similarly, the coefficient of x^5 in the binomial expansion of (x+2)^8 is 448, found by using the combination formula (8 choose 5) and taking into account the additional term of 2^3.
  • #1
mr_coffee
1,629
1
Hello everyone. I was wondering if there was a fast way to figure out the following:

THe question is:
What is the cofficient of u^16v^2 in the binomial expansion of (u^2 + v)^10?

Well the answer is 45.

I know what the binomail theorem is:
col_alg_tut54binomialthe.gif


and I can put (u^2 + v)^10 in that form
a = u^2
b = v
n = 10

(10 choose 0)*(u^2)^10 + (10 choose 1)*(u^2)^(9)*v + (10 choose 2)*(u^2)^(8)*v^2...

Okay so i found (10 choose 2) is where the coefficent of the binomial expansion of (u^2 + v)^10. But is this how your supppose to do it? Write it out like that? Or is there a faster way?

I also got 45 by just taking (10 choose 2) = 45.

So my question is, is there another way to compute this? It looks like the professor did somthing like:

(u^16)*v^2 = (u^2)^8*v^2 I see they equal each other but I'm not sure how this is connecting things together and how its figuring out the co-efficent.

ANy help would be great!
 

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  • #2
In your binomial expansion, b = v, and a = u^2.

In the same term, a is raised to the (n-2) power. (u^2)^(10-2) = (u^2)^8 = u^16, so this checks out okay. It's the right coefficient.

The coefficient with b raised to the second power is the one with (n choose 2) in front. The coefficient of this term, (10 choose 2), is 45.

- Warren
 
  • #3
Thanks Warren, I see your explanation for making sure it checks out but how did u know that the coefficient with b raised to the second power was the correct choice with (n choose 2) without writing it out? Or was that from inspection?

It looks like to find the co-effient you need to recongize 2 things,
#1. That n is whatever binomial expansion he is asking for, in this case 10, from (u^2+v)^10

#2. and look what v is raised too, which is u^16v^2

Then just toss the (10 choose 2) = 45 like you said.

Am i getting that right or did i just make an assumption its always going to work like that?

Like for instance, here is another problem:

What is the coefficient of x^5 in the binomial expansion of (x+2)^8

His answer is 448,

would it be (8 choose 5)? that gives me 56 which isn't right so i guess not, hm.
 
  • #4
Something that might help is to see all the coefficients at once a few times. Do a quick search for Pascal's triangle (use google or whatever.) Incidentally, while others are looking at this, can anyone name some realistic applications where you'd only need the coefficient of one of the terms in a binomial expansion?

edit: Ohhh, I see where you had trouble...
Don't forget that not only do get the coefficient from the combination, but when you have x^5, you also have a 2^3. That term is 56x^5 * 8
 
Last edited:
  • #5
mr_coffee said:
What is the coefficient of x^5 in the binomial expansion of (x+2)^8

His answer is 448,

would it be (8 choose 5)? that gives me 56 which isn't right so i guess not, hm.

it is x +2 in there. The 2 is important. You can't forget it. You did forget it. The coefficient of x^5y^3 in (x+y)^8 would just be 8 choose 5 (or 8 choose 3 since those are the same), as would the coefficient of x^5 in (1+x)^8
 

Related to Calculating Coefficient of u^16v^2 in (u^2 + v)^10 Expansion

What is the purpose of calculating the coefficient of u^16v^2 in (u^2 + v)^10 expansion?

The purpose of calculating the coefficient of u^16v^2 in (u^2 + v)^10 expansion is to determine the number of ways in which u^16v^2 can be obtained as a term in the expansion. This coefficient can also be used to find the probability of obtaining u^16v^2 when randomly selecting terms from the expansion.

How do you calculate the coefficient of u^16v^2 in (u^2 + v)^10 expansion?

To calculate the coefficient of u^16v^2 in (u^2 + v)^10 expansion, you can use the binomial theorem or the combination formula. The coefficient can be found by taking the product of the binomial coefficients of u^16 and v^2, which are (10 choose 16) and (10-16 choose 2), respectively. This can also be represented as (10 choose 16) * (10-16 choose 2) = 120 * 1 = 120.

What is the significance of the coefficient of u^16v^2 in (u^2 + v)^10 expansion?

The coefficient of u^16v^2 in (u^2 + v)^10 expansion represents the number of ways in which u^16v^2 can be obtained as a term in the expansion. It can also be used to find the probability of obtaining u^16v^2 when randomly selecting terms from the expansion. This coefficient is important in combinatorics and probability calculations.

How does the exponent of u and v affect the coefficient in the expansion?

The exponent of u and v determines the number of occurrences of u and v in the term. In this case, the exponent of u is 16 and the exponent of v is 2, which means that the term u^16v^2 will appear 16 times in the expansion. The coefficient is affected by the exponents because it is calculated by taking the product of the binomial coefficients of u and v.

Can the coefficient of u^16v^2 in (u^2 + v)^10 expansion be negative?

No, the coefficient of u^16v^2 in (u^2 + v)^10 expansion cannot be negative. The binomial coefficients and the combination formula only produce positive values. However, the coefficient can be zero if the exponents of u and v do not match the exponents in the term being searched for.

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