3-Find-the-term-independ-ent-of-x-in-the-expansion-of-x-1-x-2-x-1-x-12-

Question Number 27884 by das47955@mail.com last updated on 16/Jan/18

(3) Find the term independ -

ent of x in the expansion of

{(x + \frac{1}{x})}^{2} {(x - \frac{1}{x})}^{12}

Answered by Rasheed.Sindhi last updated on 16/Jan/18

{(x + \frac{1}{x})}^{2} {(x - \frac{1}{x})}^{12}

(x^{2} + 2 + \frac{1}{x^{2}}) {(x - \frac{1}{x})}^{12}

- - - -

T_{r + 1} = (\begin{matrix} n \\ r \end{matrix}) a^{n - r} b^{r}

- - - - - -

T_{r + 1} of {(x - \frac{1}{x})}^{12}

T_{r + 1} = (\begin{matrix} 12 \\ r \end{matrix}) {(x)}^{12 - r} {(- \frac{1}{x})}^{r}

T_{r + 1} = {(- 1)}^{r} (\begin{matrix} 12 \\ r \end{matrix}) {(x)}^{12 - 2 r}

T_{r + 1} multiplied by (x^{2} + 2 + \frac{1}{x^{2}})

(x^{2} + 2 + \frac{1}{x^{2}}) . T_{f + 1} = x^{2} T_{r + 1} + {2 T}_{r + 1} + (\frac{1}{x^{2}}) T_{r + 1}

= x^{2} {(- 1)}^{r} (\begin{matrix} 12 \\ r \end{matrix}) {(x)}^{12 - 2 r}

+ 2 {(- 1)}^{r} (\begin{matrix} 12 \\ r \end{matrix}) {(x)}^{12 - 2 r}

+ (\frac{1}{x^{2}}) {(- 1)}^{r} (\begin{matrix} 12 \\ r \end{matrix}) {(x)}^{12 - 2 r}

= {(- r)}^{r} (\begin{matrix} 12 \\ r \end{matrix}) {(x)}^{14 - 2 r}

+ 2 {(- 1)}^{r} (\begin{matrix} 12 \\ r \end{matrix}) {(x)}^{12 - 2 r}

+ {(- 1)}^{r} (\begin{matrix} 12 \\ r \end{matrix}) {(x)}^{10 - 2 r}

When 14 - 2 r = 0 \Rightarrow r = 7

Free of x term 1 st part of the above

{(- 1)}^{7} (\begin{matrix} 12 \\ r \end{matrix}) {(x)}^{14 - 2 (7)} = - (\begin{matrix} 12 \\ 7 \end{matrix}) . . (i)

When 12 - 2 r = 0 \Rightarrow r = 6

Free of x, 2 nd part of above

2 {(- 1)}^{6} (\begin{matrix} 12 \\ r \end{matrix}) {(x)}^{12 - 2 r} = 2 (\begin{matrix} 12 \\ 6 \end{matrix}) \dots (ii)

When 10 - 2 r = 0 \Rightarrow r = 5

Free of x, 3 rd part of the above

{(- 1)}^{7} (\begin{matrix} 12 \\ r \end{matrix}) {(x)}^{10 - 2 r} = - (\begin{matrix} 12 \\ 5 \end{matrix}) \dots (iii)

Free of x term = (i) + (ii) + (iii)

= - (\begin{matrix} 12 \\ 7 \end{matrix}) + 2 (\begin{matrix} 12 \\ 6 \end{matrix}) - (\begin{matrix} 12 \\ 5 \end{matrix})

= - 792 + 2 \times 924 - 792 = 264

Commented by Rasheed.Sindhi last updated on 16/Jan/18

Corrected now .

Answered by mrW2 last updated on 16/Jan/18

{(x + \frac{1}{x})}^{2} {(x - \frac{1}{x})}^{12}

= \frac{1}{x^{14}} {(x^{2} + 1)}^{2} {(x^{2} - 1)}^{12}

= \frac{1}{x^{14}} (x^{4} + 2 x^{2} + 1) {(x^{2} - 1)}^{12}

= \frac{1}{x^{14}} (x^{4} + 2 x^{2} + 1) \underset{k = 0}{\sum^{12}} C_{k}^{12} x^{2 k} {(- 1)}^{12 - k}

t e r m i n d e p e n d e n t o f x i s :

\frac{1}{x^{14}} \times x^{4} \times C_{5}^{12} x^{2 \times 5} {(- 1)}^{12 - 5} + \frac{1}{x^{14}} \times 2 x^{2} \times C_{6}^{12} x^{2 \times 6} {(- 1)}^{12 - 6} + \frac{1}{x^{14}} \times 1 \times C_{7}^{12} x^{2 \times 7} {(- 1)}^{12 - 7}

= - C_{5}^{12} + 2 C_{6}^{12} - C_{7}^{12}

= - 792 + 2 \times 924 - 792

= 264

Commented by mrW2 last updated on 16/Jan/18

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