SOLUTION: In the expansion of (1+x)n, two times the coefficient of x5 is equal to the sum of the coefficient of x4 + x6. Find the possible values of n.
2(nC5) = nC4 + nC6 , find n
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-> SOLUTION: In the expansion of (1+x)n, two times the coefficient of x5 is equal to the sum of the coefficient of x4 + x6. Find the possible values of n.
2(nC5) = nC4 + nC6 , find n
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Question 1040101: In the expansion of (1+x)n, two times the coefficient of x5 is equal to the sum of the coefficient of x4 + x6. Find the possible values of n.
2(nC5) = nC4 + nC6 , find n Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! 2*(n)(n-1)(n-2)(n-3)(n-4)/5! (the n-5)! cancels the (n-5)! on the bottom
That equals n(n-1)(n-2)(n-3)/4! +n(n-1)(n-2)(n-3(n-4)(n-5)/6!
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I can cancel n(n-1)(n-2)(n-3) from everything, because it is in all terms.
2(n-4)/120=1/24 + (n-4)(n-5)/720
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The first term will be (n-4)/60
Now put everything over a common denominator of 720
12(n-4)=30+(n-4)(n-5), and since they are all over the same number 720, I can remove it without changing the result.
12n-48=30+n^2-9n+20
0=n^2-21n+98
0=(n-14)(n-7)
n=14 or 7
14C5=2002
14C4=1001
14C6=3003
7C5=21
7C4=35
7C6=7
