Sy123
This too shall pass
- Joined
- Nov 6, 2011
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- HSC
- 2013
Re: HSC 2015 3U Marathon
http://www.wolframalpha.com/input/?i=(x+1/x)^3+(1+x)^5
![](https://latex.codecogs.com/png.latex?\bg_white \left(x + \frac{1}{x} \right)^3 (1+x)^5 = \frac{1}{x^3} (1+x^2)^3(1+x)^5 )
![](https://latex.codecogs.com/png.latex?\bg_white \\ $So, instead of finding the coefficient of$ \ x^2 \ $we can find the coefficient of$ \ x^5 \ $in$ \\ (1+x^2)^3 (1+x)^5 \ $since it becomes divided by the$ \ x^3 \ $in front$ )
![](https://latex.codecogs.com/png.latex?\bg_white \\ $Coefficeint of $\ x^5 \ $in$ \ \underbrace{(1+x^2)^3}_{A} \underbrace{(1+x)^5}_B \ $can be gotten by $ \\ $(1): multiplying a constant from$ \ A \ $and a$ \ x^5 \ $in B$ \\ \\ $There are no odd power terms in$ \ A \ $so the next case is$ \\ $(2): multiplying a$ \ x^2 \ $from$ \ A \ $and a$ \ x^3 \ $from B.$ \\ $(3): Multiplying a$ \ x^4 \ $from$ \ A \ $and a$ \ x \ $from B$ )
![](https://latex.codecogs.com/png.latex?\bg_white \\ $Therefore the coefficient is$ \ \binom{3}{0} \cdot \binom{5}{5} x^5 + \binom{3}{1} (x^2) \cdot \binom{5}{3} x^3 + \binom{3}{2} (x^2)^2 \cdot \binom{5}{1} x \\ \\ = x^5 \left(\binom{3}{0} \binom{5}{5} + \binom{3}{1} \binom{5}{3} + \binom{3}{2} \binom{5}{1} \right) = 46x^2 \\ \\ $Therefore the coefficient is$ \ 46 )
(as per the expansion)
Neither of those are correct:Find the coefficient of x^2 in (x+(1/x))^3 (x+1)^5
i get an answer of 55 but the solution says 92.0
http://www.wolframalpha.com/input/?i=(x+1/x)^3+(1+x)^5
![](/proxy.php?image=http%3A%2F%2Fi.imgur.com%2FFCbyz4H.png&hash=980deb38ced90f9fd71f29fdfef057a3)
(as per the expansion)
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