binomial question (1 Viewer)

fuckit1991 · Mar 4, 2009

hey guys was wondering if anyone could help me out.

i have the binomial (3x+2)^19. The question asks me to find the values of k for which x^k and x^(k+1) are equal. Answer =11

Thanks in advance

Trebla · Mar 4, 2009

I think you mean co-efficients must be equal?
$(3x+2)^{19} = \displaystyle\sum_{k=0}^{19}\binom{19}{k}2^{19-k}3^kx^k\\$We require co-efficient of $x^k$ equal to co-efficient of $x^{k+1}\\$\displaystyle\binom{19}{k}2^{19-k}3^k=\binom{19}{k+1}2^{19-(k+1)}3^{k+1}\\\Rightarrow \displaystyle\binom{19}{k}=\binom{19}{k+1}2^{-1}3\\\Rightarrow \dfrac{19!}{k!(19-k)!} = \dfrac{19!}{(k+1)!(19-(k+1))!}\dfrac{3}{2}\\\Rightarrow \dfrac{1}{19-k}=\dfrac{1}{k+1}\left(\dfrac{3}{2}\right)\\\Rightarrow 2k+2 = 57-3k\\\Rightarrow 5k=55\\\Rightarrow k=11$

binomial question (1 Viewer)

fuckit1991

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Trebla

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