Q_1 . find the term in x^6 in the expansion of (x^2 +(2/x))^9 Q_2 . in the binomial expansion of (x+(k/x))^6 , the term independent of x is 160 find the value of k. Q_3 . find the constand term in the binomial of (x+(3/x))^(12) .


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Question Number 35357 by Rio Mike last updated on 18/May/18

$Q_{1} . f i n d t h e t e r m i n x^{6} i n t h e$ $e x p a n s i o n o f {(x^{2} + \frac{2}{x})}^{9}$ $Q_{2} . i n t h e b i n o m i a l e x p a n s i o n o f$ ${(x + \frac{k}{x})}^{6}, t h e t e r m i n d e p e n d e n t o f x$ $i s 160 f i n d t h e v a l u e o f k .$ $Q_{3} . f i n d t h e c o n s t a n d t e r m i n t h e$ $b i n o m i a l o f {(x + \frac{3}{x})}^{12} .$
	Answered by MJS last updated on 18/May/18

	${(a + b)}^{n} = \underset{k = 0}{\sum^{n}} (\begin{matrix} n \\ k \end{matrix}) \times a^{n - k} b^{k}$ ${(x^{p} + \frac{r}{x^{q}})}^{n} = \frac{1}{x^{n q}} {(x^{p + q} + r)}^{n} =$ $= \frac{1}{x^{n q}} \times \underset{k = 0}{\sum^{n}} (\begin{matrix} n \\ k \end{matrix}) \times r^{k} \times {(x^{p + q})}^{n - k} =$ $= \frac{1}{x^{n q}} \times \underset{k = 0}{\sum^{n}} (\begin{matrix} n \\ k \end{matrix}) \times r^{k} \times x^{(p + q) (n - k)} =$ $= \underset{k = 0}{\sum^{n}} (\begin{matrix} n \\ k \end{matrix}) \times r^{k} \times x^{n p - k (p + q)}$ ${(x^{2} + \frac{2}{x})}^{9} =$ $[n = 9; p = 2; q = 1; r = 2]$ $= \underset{k = 0}{\sum^{9}} (\begin{matrix} 9 \\ k \end{matrix}) \times 2^{k} x^{18 - 3 k}$ $18 - 3 k = 6 \Rightarrow k = 4$ $(\begin{matrix} 9 \\ 4 \end{matrix}) \times 2^{4} = \frac{9!}{4! \times 5!} \times 16 = 2016$ ${(x + \frac{k}{x})}^{6} let k = r to avoid confusion$ ${(x + \frac{r}{x})}^{6} =$ $[n = 6; p = 1; q = 1]$ $= \underset{k = 0}{\sum^{6}} (\begin{matrix} 6 \\ k \end{matrix}) \times r^{k} x^{6 - 2 k}$ $6 - 2 k = 0 \Rightarrow k = 3$ $(\begin{matrix} 6 \\ 3 \end{matrix}) \times r^{3} = 160$ $\frac{6!}{{(3!)}^{2}} r^{3} = 160$ $20 r^{3} = 160$ $r^{3} = 8$ $r = 2$ $so the searched k = 2$ ${(x + \frac{3}{x})}^{12} =$ $[n = 12; p = 1; q = 1; r = 3]$ $= \underset{k = 0}{\sum^{12}} (\begin{matrix} 12 \\ k \end{matrix}) \times 3^{k} x^{12 - 2 k}$ $12 - 2 k = 0 \Rightarrow k = 6$ $\frac{12!}{{(6!)}^{2}} 3^{6} = 924 \times 729 = 673596$
	Answered by ajfour last updated on 18/May/18

	$Q . 1)$ ${(x^{2} + \frac{2}{x})}^{9} = Σ T_{r + 1} = \underset{r = 0}{\sum^{9}}^{9} C_{r} {(x^{2})}^{r} {(\frac{2}{x})}^{9 - r}$ $f o r T_{r + 1} t o h a v e x^{6},$ $2 r - 9 + r = 6 \Rightarrow r = 5$ $c o e f f i c i e n t =^{9} C_{5} \times 2^{4} = 126 \times 16$ $= 2016 .$ $Q . 2)$ ${(x + \frac{k}{x})}^{6} = \underset{r = 0}{\sum^{6}} T_{r + 1} = \underset{r = 0}{\sum^{6}}^{6} C_{r} x^{r} {(\frac{k}{x})}^{6 - k}$ $f o r t e r m i n d e p e n d e n t o f x$ $r = 6 - r \Rightarrow r = 3$ $c o e f f . o f t e r m w i t h x^{0} =^{6} C_{3} k^{3} = 160$ $\Rightarrow 20 k^{3} = 160$ $o r k = 2$ $Q . 3)$ ${(x + \frac{3}{x})}^{12} = \underset{r = 0}{\sum^{12}} T_{r + 1} = \underset{r = 0}{\sum^{12}}^{12} C_{r} x^{r} {(\frac{3}{x})}^{12 - r}$ $f o r t e r m w i t h x^{0},$ $r + 12 - r = 0 \Rightarrow r = 6$ $t h e t e r m =^{12} C_{6} (3^{6})$ $= \frac{7 \times 8 \times 9 \times 10 \times 11 \times 12}{6 \times 5 \times 4 \times 3 \times 2} \times 81 \times 9$ $= \frac{7 \times 8 \times 9 \times 11}{6} \times 81 \times 9$ $= 84 \times 11 \times 9 \times 81$ $= 673596 .$
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