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# Write down and simplifyThe ${\text{1}}{{\text{4}}^{th}}$term of ${\left( {{2^{10}} - {2^7}x} \right)^{\dfrac{{13}}{2}}}$

Last updated date: 31st Mar 2023
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Hint: Use general term $\left( {{\text{i}}{\text{.e}}{\text{.}}{{\left( {r + 1} \right)}^{th}}{\text{ term}}} \right)$ in the expansion of ${\left( {1 + x} \right)^n}$=
${{\text{T}}_{r + 1}} = \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)............\left( {n - r + 1} \right)}}{{r!}}{x^r}$

Given equation is:
${\left( {{2^{10}} - {2^7}x} \right)^{\dfrac{{13}}{2}}}$
Take ${2^{10}}$as common
$\Rightarrow {\left( {{2^{10}} - {2^7}x} \right)^{\dfrac{{13}}{2}}} = {\left( {{2^{10}}} \right)^{\dfrac{{13}}{2}}}{\left( {1 - \dfrac{{{2^7}}}{{{2^{10}}}}x} \right)^{\dfrac{{13}}{2}}} = {2^{65}}{\left( {1 - \dfrac{x}{8}} \right)^{\dfrac{{13}}{2}}}$
Now as we know the general term $\left( {{\text{i}}{\text{.e}}{\text{.}}{{\left( {r + 1} \right)}^{th}}{\text{ term}}} \right)$ in the expansion of ${\left( {1 + x} \right)^n}$is given as
${{\text{T}}_{r + 1}} = \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)............\left( {n - r + 1} \right)}}{{r!}}{x^r}$
$\therefore {14^{th}}$Term is ${T_{14}}$, for $r = 13$
$\therefore {{\text{T}}_{14}} = \dfrac{{{2^{65}}}}{{13!}}\left\{ \dfrac{{13}}{2}\left( {\dfrac{{13}}{2} - 1} \right)\left( {\dfrac{{13}}{2} - 2} \right)\left( {\dfrac{{13}}{2} - 3} \right)\left( {\dfrac{{13}}{2} - 4} \right)\left( {\dfrac{{13}}{2} - 5} \right)\left( {\dfrac{{13}}{2} - 6} \right) \\ {\text{ }} \times \left( {\dfrac{{13}}{2} - 7} \right)\left( {\dfrac{{13}}{2} - 8} \right)\left( {\dfrac{{13}}{2} - 9} \right)\left( {\dfrac{{13}}{2} - 10} \right)\left( {\dfrac{{13}}{2} - 11} \right)\left( {\dfrac{{13}}{2} - 12} \right){\left( { - \dfrac{x}{8}} \right)^{13}} \\ \right\}$
$\therefore {{\text{T}}_{14}} \Rightarrow {2^{65}}\left[ {\dfrac{{\dfrac{{13}}{2}\left( {\dfrac{{11}}{2}} \right)\left( {\dfrac{9}{2}} \right)\left( {\dfrac{7}{2}} \right)\left( {\dfrac{5}{2}} \right)\left( {\dfrac{3}{2}} \right)\left( {\dfrac{1}{2}} \right)\left( { - \dfrac{1}{2}} \right)\left( { - \dfrac{3}{2}} \right)\left( { - \dfrac{5}{2}} \right)\left( { - \dfrac{7}{2}} \right)\left( { - \dfrac{9}{2}} \right)\left( { - \dfrac{{11}}{2}} \right)}}{{13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{\left( { - \dfrac{x}{8}} \right)}^{13}}} \right]$
$\Rightarrow {2^{65}}\left[ {\dfrac{{13.11.9.7.5.3.1.1.3.5.7.9.11}}{{{2^{13}}{{.13.12.11.10.9.8.7.6.5.4.3.2.1.8}^{13}}}}\left( { - {x^{13}}} \right)} \right]$
$\Rightarrow {2^{13}}\left[ {\dfrac{{3.5.7.9.11}}{{12.10.8.6.4.2}}} \right]\left( { - {x^{13}}} \right) = - 1848{x^{13}}$
So, this is the required value of the ${14^{th}}$ term.

Note: - In such types of questions the key concept is that we have to remember the general term $\left( {{\text{i}}{\text{.e}}{\text{.}}{{\left( {r + 1} \right)}^{th}}{\text{ term}}} \right)$ which is stated above in the expansion of ${\left( {1 + x} \right)^n}$, then calculate the required term using this formula and simplify then we will get the required answer.