Annie M.

asked • 04/14/15

Find the coefficient of x^6 in the expansion of (2x+1)^12

Find the coefficient of x^6 in the expansion of (2x+1)^12

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Andrew M. answered • 04/15/15

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If we use Pascal's Triangle to find the coefficients of a polynomial of form (a+b)n we get the 12th iteration as being
 
a12+12a11b + 66a10b2 + 220a9b3 + 495a8b4 + 792a7b5 + 924a6b6 + 792a5b7 + 495a4b8 + 220a3b9 + 66a2b10 + 12ab11 + b12
 
In this particular equation of (2x+1)12 the "a" is 2x and the "b" term is 1...
 
The x6 term in the expansion comes from 924a6b6 so replace a with 2x and replace b with 1
 
So we have 924(2x)6(1)6 = 924(2)6x6(1) = 924(64)x6 = 59136x6
 
Thus, the coefficient of the x6 term is 59,136
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Andrew M.

Another way to do this same problem is to use the formula to find the (k+1)st term in a polynomial (a+b)n
 
Term (k+1) = [n!/(k!(n-k)!)]an-kbk
 
This is a factorial equation
 
The x6 term in the expansion of (a+b)12 will be the 7th term in the sequence because the exponents on the
term will start at a12 and decrease by a power of 1 with each subsequent term so we get powers of the 1st
term going from a12, a11, a10,...
 
So for the 7th term in the expansion k=6
n = 12, a = 2x and b = 1
 
Plugging that into our equation we get    [(12!/6!(12-6)!)][ (2x)12-6(1)6]
 
Working this out we get [(12)(11)(10)(9)(8)(7)(6!)]/[(6!)(6!)] [(2x)6(1)]
 
We can cancel a 6! from top and bottom leaving
 
[[(12)(11)(10)(9)(8)(7)]/6!]26x6
 
= (924)(64)x6
 
= 59136x6
 
 
 
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04/15/15

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