Let ${\left( {2{x^2} + 3x + 4} \right)^{10}} = \sum\limits_{r = 0}^{20} {{a_r}{x^r}}$
Then ${{{a_7}} \over {{a_{13}}}}$ is equal to ______.
Answer (integer)
8
Solution
<b>Note : </b> <b>Multinomial Theorem : </b>
<br><br>The general term of ${\left( {{x_1} + {x_2} + ... + {x_n}} \right)^n}$ the expansion is
<br><br>${{n!} \over {{n_1}!{n_2}!...{n_n}!}}x_1^{{n_1}}x_2^{{n_2}}...x_n^{{n_n}}$
<br><br>where n<sub>1</sub> + n<sub>2</sub> + ..... + n<sub>n</sub> = n
<br><br>Here, in ${(2{x^2} + 3x + 4)^{10}}$ general term is <br><br>$$ = {{10!} \over {{n_1}!{n_2}!{n_3}!}}{(2{x^2})^{{n_1}}}{(3x)^{{n_2}}}{(4)^{{n_3}}}$$<br><br>$$ = {{10!} \over {{n_1}!{n_2}!{n_3}!}}{.2^{{n_1}}}{.3^{{n_2}}}{.4^{{n_3}}}.{x^{2{n_1} + {n_2}}}$$<br><br>$\therefore$ Coefficient of ${x^{2{n_1} + {n_2}}}$ is <br><br>${{10!} \over {{n_1}!{n_2}!{n_3}!}}{.2^{{n_1}}}{.3^{{n_2}}}{.4^{{n_3}}}$<br><br>where ${n_1} + {n_2} + {n_3} = 10$<br><br> For, Coefficient of x<sup>7</sup> : <br>2n<sub>1</sub> + n<sub>2</sub> = 7<br><br>Possible values of n<sub>1</sub>, n<sub>2</sub> and n<sub>3</sub> are <br><br><table>
<thead>
<tr>
<th>${n_1}$</th>
<th>${n_2}$</th>
<th>${n_3}$</th>
</tr>
</thead>
<tbody>
<tr>
<td>3</td>
<td>1</td>
<td>6</td>
</tr>
<tr>
<td>2</td>
<td>3</td>
<td>5</td>
</tr>
<tr>
<td>1</td>
<td>5</td>
<td>4</td>
</tr>
<tr>
<td>0</td>
<td>7</td>
<td>3</td>
</tr>
</tbody>
</table><br><br>$\therefore$ Coefficient of x<sup>7</sup><br><br>$$ = {{10!} \over {3!1!6!}}{(2)^3}{(3)^1}{(4)^6} + {{10!} \over {3!1!6!}}{(2)^2}{(3)^3}{(4)^5} + {{10!} \over {1!5!4!}}{(2)^1}{(3)^5}{(4)^4} + {{10!} \over {0!7!3!}}{(2)^0}{(3)^7}{(4)^3}$$<br><br>Coefficient of x<sup>13</sup> = a<sub>13</sub><br><br>Here 2n<sub>1</sub> + n<sub>2</sub> = 13<br><br><br>possible values of n<sub>1</sub>, n<sub>2</sub> and n<sub>3</sub> are<br><br><table>
<thead>
<tr>
<th>${n_1}$</th>
<th>${n_2}$</th>
<th>${n_3}$</th>
</tr>
</thead>
<tbody>
<tr>
<td>6</td>
<td>1</td>
<td>3</td>
</tr>
<tr>
<td>5</td>
<td>3</td>
<td>2</td>
</tr>
<tr>
<td>4</td>
<td>5</td>
<td>1</td>
</tr>
<tr>
<td>3</td>
<td>7</td>
<td>0</td>
</tr>
</tbody>
</table><br><br>$\therefore$ Coefficient of x<sup>13</sup><br><br>$$ = {{10!} \over {6!1!3!}}{(2)^6}{(3)^1}{(4)^3} + {{10!} \over {5!3!2!}}{(2)^5}{(3)^3}{(4)^2} + {{10!} \over {4!5!1!}}{(2)^4}{(3)^5}{(4)^1} + {{10!} \over {3!7!0!}}{(2)^3}{(3)^7}{(4)^0}$$<br><br>$\therefore$ ${{{a_7}} \over {{a_{13}}}} = 8$
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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