xxmariposa4xx
Answered

Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Ask your questions and receive accurate answers from professionals with extensive experience in various fields on our platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

Stuck on this problem

Stuck On This Problem class=

Sagot :

sqdancefan

9514 1404 393

Answer:

  -8,257,536·u^5·v^10

Step-by-step explanation:

The expansion of (a +b)^n is ...

  (c0)a^nb^0 +(c1)a^(n-1)b^1 +(c2)a^(n-2)b^2 +... +(ck)a^(n-k)b^k +... +(cn)a^0b^n

Then the k-th term is (ck)a^(n-k)b^k, where k is counted from 0 to n.

The value of ck can be found using Pascal's triangle, or by the formula ...

  ck = n!/(k!(n-k)!) . . . . where x! is the factorial of x, the product of all positive integers less than or equal to x.

This expansion has 11 terms, so the middle one is the one for k=5. That term will be ...

  5th term = (10!/(5!(10-5)!)(2u)^(10-5)(-4v^2)^5

  = (252)(32u^5)(-1024v^10) = -8,257,536·u^5·v^10

Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.