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

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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