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Binomial theorem question (1 Viewer)

Thread starter infatuated-x3
Start date Apr 8, 2010

infatuated-x3

Procrastinator

#1

Find the value of k if the coefficients of x^k and x^(k+1) in (3x+2)^19 are equal.

Answer is 11 but I have no idea how to do the working out.

D

Drongoski

Well-Known Member

#2

$\100dpi (3x+2)^{19}\ \ =\ \ \sum_{i=0}^{19}\ T_{i}\ \ = \sum_{i=0}^{19}\binom{19}{i}(3x)^{i}\cdot 2^{19-i}\\ \\ Now\ \ coeff\ of\ x^{k}\ in\ T_{k} \ \ =\ \ coeff\ of\ x^{k+1}\ in\ T_{k+1}\\ \\ \therefore \ \ \binom{19}{k}\cdot 3^{k}\cdot 2^{19-k}\ \ =\ \ \binom{19}{k+1}\cdot 3^{k+1}\cdot 2^{19-(k+1)}\\ \\ \therefore \ \ \binom{19}{k}\cdot 2\ \ =\ \ \binom{19}{k+1}\cdot 3\\ \\ i.e.\ \ \frac{19!}{k!(19-k)!}\cdot 2\ \ =\ \ \frac{19!}{(k+1)!(19-k-1)!}\cdot 3\ \\ \\ \therefore \ \ \frac{2}{19-k}\ \ =\ \ \frac{3}{k+1}\\ \\ solving\ gives\ \ k=11$

Last edited: Apr 8, 2010

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