Find th greatest coefficients in the expansion of (3a+5b)^(18) 2)If three consecutive coefficient of (1+x)^n are 28,56,70. find the value of n


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Question Number 75242 by peter frank last updated on 08/Dec/19

$F i n d t h g r e a t e s t c o e f f i c i e n t s$ $i n t h e e x p a n s i o n o f$ ${(3 a + 5 b)}^{18}$ $2) I f t h r e e c o n s e c u t i v e$ $c o e f f i c i e n t o f {(1 + x)}^{n} a r e 28, 56, 70 .$ $f i n d t h e v a l u e o f n$
	Answered by mr W last updated on 08/Dec/19

	$(1)$ ${(3 a + 5 b)}^{18} = \underset{k = 0}{\sum^{18}} C_{k}^{18} {(3 a)}^{k} {(5 b)}^{18 - k}$ $= \underset{k = 0}{\sum^{18}} 3^{k} 5^{18 - k} C_{k}^{18} a^{k} b^{18 - k}$ $a_{k} = 3^{k} 5^{18 - k} C_{k}^{18}$ $a_{k + 1} = 3^{k + 1} 5^{17 - k} C_{k + 1}^{18}$ $\frac{a_{k + 1}}{a_{k}} = \frac{3^{k + 1} 5^{17 - k} C_{k + 1}^{18}}{3^{k} 5^{18 - k} C_{k}^{18}} = \frac{3 (18 - k)}{5 (k + 1)} < 1$ $5 k + 5 > 54 - 3 k$ $k > \frac{49}{8} = 6.12 = 7 \Rightarrow a_{8} < a_{7}$ $m a x . a_{k} i s a_{7} = 3^{7} 5^{11} C_{7}^{18} = 3398392968750000$
	Commented by peter frank last updated on 09/Dec/19

	$t h a n k y o u v e r y m u c h$
	Answered by mr W last updated on 08/Dec/19

	$(2)$ $C_{k}^{n} = \frac{n!}{k! (n - k)!} = 28$ $C_{k + 1}^{n} = \frac{n!}{(k + 1)! (n - k - 1)!} = 56$ $\frac{n - k}{k + 1} = \frac{56}{28} = 2$ $\Rightarrow n = 3 k + 2$ $C_{k + 2}^{n} = \frac{n!}{(k + 2)! (n - k - 2)!} = 70$ $\frac{n - k - 1}{k + 2} = \frac{70}{56}$ $\Rightarrow 28 n = 63 k + 98$ $\Rightarrow 28 (3 k + 2) = 63 k + 98$ $\Rightarrow k = 2$ $\Rightarrow n = 8$
	Commented by peter frank last updated on 09/Dec/19

	$t h a n k y o u v e r y m u c h$
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