binomial question (1 Viewer)

viraj30 · Jul 15, 2012

can someone show me how to do this following question?

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

Thanks in advance!

Sy123 · Jul 15, 2012

$\binom{19}{k} (2)^{19-k} (3x)^{k}=\binom{19}{k+1} (2)^{18-k} (3x)^{k+1} \\ \\ \frac{19!}{(19-k)!k!} \times 2 = \frac{19!}{(19-k-1)!(k+1)!} \times 3 \\ \\ (2 \times 19!)(19-k-1)!(k+1)!=(3 \times 19!)(19-k)!k! \\ \\ (2 \times 19!)(k+1)=(3 \times 19!)(19-k) \\ \\ 2k+2=57-3k \\ \\ k=11$

I skipped some cancellation steps, if you are confused about how things have disappeared, like from the first step to the second step, its because I cancelled out the 2 and 3 powers, (by dividing)

binomial question (1 Viewer)

viraj30

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Sy123

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