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Stuck on this problem

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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

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