If-the-coefficients-of-x-k-and-x-k-1-in-the-expansion-2-3x-19-are-equal-what-is-the-value-of-k-

Question Number 100587 by bobhans last updated on 27/Jun/20

If the coefficients of x^{k} and x^{k + 1} in the

expansion {(2 + 3 x)}^{19} are equal, what is

the value of k ?

Commented by bobhans last updated on 27/Jun/20

thank you both

Answered by maths mind last updated on 27/Jun/20

{(2 + 3 x)}^{19} = Σ C_{19}^{k} {(3 x)}^{k} 2^{19 - k}

\Rightarrow C_{19}^{m} 3^{m} 2^{19 - m} = C_{19}^{m + 1} 3^{m + 1} 2^{19 - m - 1} \Rightarrow

\frac{C_{19}^{m}}{C_{19}^{m + 1}} = \frac{3}{2}

\Rightarrow \frac{(m + 1)! . (18 - m)!}{m! . (19 - m)!} = \frac{3}{2} \Leftrightarrow 2 (m + 1) = 3 (19 - m)

\Rightarrow 5 m = 55 \Rightarrow m = 11

Answered by john santu last updated on 27/Jun/20

by Binomial Theorem

{(2 + 3 x)}^{19} = \underset{n = 0}{\sum^{19}} (\begin{matrix} 19 \\ n \end{matrix}) {(2)}^{19 - n} {(3 x)}^{n}

w e w a n t t o f i n d k s o t h a t

(\begin{matrix} 19 \\ k \end{matrix}) {(2)}^{19 - k} {(3 x)}^{k} = (\begin{matrix} 19 \\ k + 1 \end{matrix}) {(2)}^{18 - k} {(3 x)}^{k + 1}

\Rightarrow 2 . \frac{19!}{k! (19 - k)!} = 3 . \frac{19!}{(k + 1)! (18 - k)!}

2 . \frac{19!}{(19 - k) . k! (18 - k)!} = 3 . \frac{19!}{(k + 1) . k! (18 - k)!}

\frac{2}{19 - k} = \frac{3}{k + 1} \Leftrightarrow k = 11