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**Hint:**Binomial expansion (or Binomial Theorem) which states that ${{\left( x+y \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}{{x}^{n-r}}{{y}^{r}}}$. Here use the binomial expansion for negative exponents i.e., $(1+x)^{-n} = 1 - nx + \dfrac{n(n+1)}{2!}x^2 + \dfrac{n(n+1)(n+2)}{3!}x^3 + . . . . . $

**Complete step by step solution:**

We have the expression ${{\left( 1+\dfrac{1}{2}a \right)}^{-4}}$. We have to write its expansion upto $4$ terms. We will use the formula for binomial expansion of terms which is $(1+x)^{-n} = 1 - nx + \dfrac{n(n+1)}{2!}x^2 + \dfrac{n(n+1)(n+2)}{3!}x^3 + . . . . . $

On substituting the values that is $n=-4$ and $x=\dfrac{1}{2}a$

${{\left( 1+\dfrac{1}{2}a \right)}^{-4}} = 1 - (-4)\left(\dfrac{1}{2}a\right) + \dfrac{(-4)(-4+1)}{2!}\left(\dfrac{1}{2}a\right)^2 + \dfrac{-4(-4+1)(-4+2)}{3!}\left(\dfrac{1}{2}a\right)^3$

On simplifying the above equation, we get

${{\left( 1+\dfrac{1}{2}a \right)}^{-4}} = 1 + (2a) + {(-2)(-3)}\left(\dfrac{1}{4}a^2\right) + \dfrac{-4(-3)(-2)}{3\times 2}\left(\dfrac{1}{8}a^3\right) $

${{\left( 1+\dfrac{1}{2}a \right)}^{-4}} = 1 + (2a) + \left(\dfrac{3}{2}a^2\right) - \left(\dfrac{1}{2}a^3\right) $

Hence we get the expansion of $\left(1+\dfrac{1}{2}a\right)^{4}$ upto 4 terms as $1+2a+\dfrac{3}{2}a^2-\dfrac{1}{2}a^3$

**Note:**Binomial expansion (also known as Binomial Theorem) describes the algebraic expansion of powers of a binomial. We expand the polynomial \[{{\left( x+y \right)}^{n}}\] into a sum involving terms of the form \[a{{x}^{b}}{{y}^{c}}\], where \[b\] and \[c\] are non-negative integers with \[b+c=n\] and the coefficient \[a\] of each term is a specific positive integer. The coefficient \[a\] in the term \[a{{x}^{b}}{{y}^{c}}\] is known as the binomial coefficient \[\left( \begin{align}

& n \\

& b \\

\end{align} \right)\] or \[\left( \begin{align}

& n \\

& c \\

\end{align} \right)\]. These coefficients for varying \[n\] and \[b\] can be arranged to form a Pascal’s Triangle. While using the formula of binomial expansion, one must keep in mind that \[n\] is a non-negative integer. That’s why to expand the expression \[{{\left( 1+\dfrac{1}{2}a \right)}^{-4}}\], we wrote it in terms of fraction to get positive value of \[n\].

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