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Expert: - 5/29/2009


Question
The question asks to find the coefficient of x^6 in the
expansion of (2+x)(4-3x)^16. I manage to get to the stage
where I have worked out the expansion to be 2×16!/6!10!
×4^10×3^6- 1×16!/5!11!×4^11×3^5 but now I don't know what
to do to find the final answer. Your help would be much
appreciated thanks

Answer
The coefficient on x^6 would be the sum of two coefficients.
It would be 2*(coefficient on x^6) + (coefficient on x^5).

If you expand (a+b)^n, you get
a^n + n*a^(n-1)b^n + n(n-1)a^(n-2)b^2/2 ...
where each term is multiplied by a certain value.

What this turns out to be is C(n,k) where k is the term and n is 16
and C(n,k) = n!/(k!(n-k)!).

I will assume that this was what was done to get to
2×16!/6!10!×4^10×3^6- 1×16!/5!11!×4^11×3^5.

Part of the first term is 2*16*15*14*13*12*11/720 = 16016.
4^10 is 1,048,576.  3^6 is 729.  
Multiply by these to get the first term.
That gives a number in scientific notation around 10^13.

Part of the second term is 4368.
4^11 = 4,194,304.  3^5 = 243.
Multiply by these to get the second term.
That gives a number in scientific notation around 10^12.

Take the difference.  That gives 7.79089E+12,
which is 7,790,890,000,000, or almost 8 trillion in
the US numbering system since in the US we go first 3 are hundereds, next 3 are thousands, next 3 are millions, next 3 are billions, and next 3 are trillions.  

If I understand the European numbering system,
it is 8 billion since they got 7 billion,
(790 thousand 890) million, and 0 (thousand and hundreds).

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