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binomial problem (1 Viewer)
- Thread starter dathat
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integral95
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how did you get that 9c0 9c2 +9c0 9c1 k from that?+kx^2) = \binom{9}{0}(1+x)^9 +\binom{9}{1}(1+x)^8kx^2+\binom{9}{3}(1+x)^7(kx^2)^2+... \\ \\ \\ $Now we only focuss on the first two terms of the expansion as the other terms will have powers greater than x^2$ \\ \\ \\ $so the coefficient of$ \ \ x^2 \ \ $is only in$ \\ \\ \binom{9}{0}(1+x)^9 +\binom{9}{1}(1+x)^8kx^2 $ and you get$ \ \ \binom{9}{0}\binom{9}{2}+\binom{9}{0}\binom{9}{1}k = 0 )
From 9c0(1+x)^9, the coefficient of x^2 = 9c0*9c2 (9c2 is the coefficient of x^2 in the expansion of (1+x)^9). Similarly, the coefficient of x^2 in 9c1(1+x)^8kx^2 = 9c0 which you multiply by 9c1, giving 9co * 9c1.
integral95
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Ekman
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Are you sure that its 9c1 * 9c0 * k for the coefficient of x^2 in 9c1(1+x)^8kx^2 ? I know it doesn't matter but it should be 9C1 * 8c0 * kThe coefficient of x^2 inand the x^2 in
is
then you add those 2 terms together.
integral95
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ahhhh damn you're right,I always screw up little things like that, luckily I won't get the wrong answer.