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Hint: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:

${\left( {1 + 2x} \right)^{ - \dfrac{1}{2}}}$

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

\[{{\text{T}}_{r + 1}} = \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)............\left( {n - r + 1} \right)}}{{r!}}{x^r}\]

\[\therefore {8^{th}}\]Term is\[{{\text{T}}_8}\], for \[{\text{r = 7}}\]

\[

\therefore {{\text{T}}_8} = \left[ {\dfrac{{\dfrac{{ - 1}}{2}\left( {\dfrac{{ - 1}}{2} - 1} \right)\left( {\dfrac{{ - 1}}{2} - 2} \right)\left( {\dfrac{{ - 1}}{2} - 3} \right)\left( {\dfrac{{ - 1}}{2} - 4} \right)\left( {\dfrac{{ - 1}}{2} - 5} \right)\left( {\dfrac{{ - 1}}{2} - 6} \right)}}{{7!}}{{\left( {2x} \right)}^7}} \right] \\

\Rightarrow \left[ {\dfrac{{\dfrac{{ - 1}}{2}\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)\left( {\dfrac{{ - 13}}{2}} \right)}}{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{\left( {2x} \right)}^7}} \right] \\

\Rightarrow \left[ {\dfrac{{{{\left( { - 1} \right)}^7}.1.3.5.7.9.11.13}}{{{2^7}.7.6.5.4.3.2.1}}{2^7}.{x^7}} \right] \\

\Rightarrow - \left[ {\dfrac{{9.11.13}}{{6.4.2.1}}} \right]\left( {{x^7}} \right) = - \dfrac{{1287}}{{48}}{x^7} = - \dfrac{{429}}{{16}}{x^7} \\

\]

So, this is the required \[{8^{th}}\]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.

Given equation is:

${\left( {1 + 2x} \right)^{ - \dfrac{1}{2}}}$

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

\[{{\text{T}}_{r + 1}} = \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)............\left( {n - r + 1} \right)}}{{r!}}{x^r}\]

\[\therefore {8^{th}}\]Term is\[{{\text{T}}_8}\], for \[{\text{r = 7}}\]

\[

\therefore {{\text{T}}_8} = \left[ {\dfrac{{\dfrac{{ - 1}}{2}\left( {\dfrac{{ - 1}}{2} - 1} \right)\left( {\dfrac{{ - 1}}{2} - 2} \right)\left( {\dfrac{{ - 1}}{2} - 3} \right)\left( {\dfrac{{ - 1}}{2} - 4} \right)\left( {\dfrac{{ - 1}}{2} - 5} \right)\left( {\dfrac{{ - 1}}{2} - 6} \right)}}{{7!}}{{\left( {2x} \right)}^7}} \right] \\

\Rightarrow \left[ {\dfrac{{\dfrac{{ - 1}}{2}\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)\left( {\dfrac{{ - 13}}{2}} \right)}}{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{\left( {2x} \right)}^7}} \right] \\

\Rightarrow \left[ {\dfrac{{{{\left( { - 1} \right)}^7}.1.3.5.7.9.11.13}}{{{2^7}.7.6.5.4.3.2.1}}{2^7}.{x^7}} \right] \\

\Rightarrow - \left[ {\dfrac{{9.11.13}}{{6.4.2.1}}} \right]\left( {{x^7}} \right) = - \dfrac{{1287}}{{48}}{x^7} = - \dfrac{{429}}{{16}}{x^7} \\

\]

So, this is the required \[{8^{th}}\]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.

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