Q1. a) solve for x 9^x +5(3^x )=6 b)write down the first 4 terms in the binomial expansion of (1−3x)^7 c)the sum S_n of the first n^(th) terms is given by S_(n ) = 3(1−((2/3))^n ) find d) the common ratio e) the sum to infinity of the progression


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Question Number 35304 by Rio Mike last updated on 17/May/18

$Q 1 . a) s o l v e f o r x 9^{x} + 5 (3^{x}) = 6$ $b) w r i t e d o w n t h e f i r s t 4 t e r m s$ $i n t h e b i n o m i a l e x p a n s i o n o f {(1 - 3 x)}^{7}$ $c) t h e s u m S_{n} o f t h e f i r s t n^{t h} t e r m s$ $i s g i v e n b y S_{n} = 3 (1 - {(\frac{2}{3})}^{n}) f i n d$ $d) t h e c o m m o n r a t i o$ $e) t h e s u m t o i n f i n i t y o f t h e p r o g r e s s i o n$
Commented by prakash jain last updated on 17/May/18

$9^{x} + 5 \cdot 3^{x} = 6$ $3^{2 x} + 5 \cdot 3^{x} - 6 = 0$ $3^{2 x} + 6 \cdot 3^{x} - 3^{x} - 6 = 0$ $3^{x} (3^{x} + 6) - (3^{x} + 6) = 0$ $(3^{x} - 1) (3^{x} + 6) = 0$ $3^{x} = - 6 (n o t p o s s i b l e f o f x \in R$ $3^{x} = 1 \Rightarrow x = 0$ $a n s x = 0$
Commented by Rasheed.Sindhi last updated on 18/May/18

$Welcome back SIR! I {haven}^{'} t seen your$ $post for long time . Are you too busy$ $nowadays ?$
Commented by rahul 19 last updated on 18/May/18
No problem sir ��
Commented by prakash jain last updated on 18/May/18

$Was very busy in office . Could$ $only read the post once in a week .$
	Answered by MJS last updated on 18/May/18

	${(3^{x})}^{2} + 5 (3^{x}) - 6 = 0$ $(3^{x}) = - \frac{5}{2} \pm \sqrt{\frac{25}{4} + 6} = - \frac{5}{2} \pm \frac{7}{2}$ $(3^{x}) = - 6 \lor (3^{x}) = 1$ $x_{2} = 0$ $x_{1} = \frac{\ln 6}{\ln 3} + \frac{(2 n + 1) π}{\ln 3} i; n \in N_{0}$ ${(a - b)}^{7} = (\begin{matrix} 7 \\ 0 \end{matrix}) \times a^{7} - (\begin{matrix} 7 \\ 1 \end{matrix}) \times a^{6} b + (\begin{matrix} 7 \\ 2 \end{matrix}) \times a^{5} b^{2} - (\begin{matrix} 7 \\ 3 \end{matrix}) \times a^{4} b^{3} + (\begin{matrix} 7 \\ 4 \end{matrix}) \times a^{3} b^{4} - (\begin{matrix} 7 \\ 5 \end{matrix}) \times a^{2} b^{5} + (\begin{matrix} 7 \\ 6 \end{matrix}) \times {a b}^{6} - (\begin{matrix} 7 \\ 7 \end{matrix}) \times {i b}^{7}$ ${(1 - 3 x)}^{7} = 1 - 21 x + 189 x^{2} - 945 x^{3} + 2835 x^{4} - 5103 x^{5} + 5103 x^{6} - 2187 x^{7}$ $the ratio is \frac{2}{3}$ $p_{n} = {(\frac{2}{3})}^{n}$ $p_{n} = q^{n} \Rightarrow S_{n} = \frac{q^{n + 1} - 1}{q - 1}$ $S_{\infty} = \frac{1}{1 - q} = 3$
	Answered by Rasheed.Sindhi last updated on 27/May/18

	$(c) S_{n} = 3 (1 - {(\frac{2}{3})}^{n})$ $First term = a = 3 (1 - {(\frac{2}{3})}^{1}) = 1$ $⋇ Common ratio r$ $1 + r = \frac{S_{2}}{S_{1}} = \frac{3 (1 - {(\frac{2}{3})}^{2})}{3 (1 - {(\frac{2}{3})}^{1})} [\frac{S_{2}}{S_{1}} = \frac{a + a r}{a} = 1 + r]$ $= \frac{(1 - \frac{2}{3}) (1 + \frac{2}{3})}{(1 - \frac{2}{3})} = \frac{5}{3}$ $r = \frac{5}{3} - 1 = \frac{2}{3}$ $⋇ Sum of infinite terms = \frac{a}{1 - r} = \frac{1}{1 - \frac{2}{3}}$ $= \frac{1}{1 / 3} = 3$
	Commented by Rasheed.Sindhi last updated on 27/May/18

	$Corrected .$
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