Expand to 4 terms the following expressions: \[{{\left( 1+\dfrac{1}{2}a \right)}^{-4}}\]

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Hint: Use the formula for binomial expansion (or Binomial Theorem) which states that \[{{\left( x+y \right)}^{n}}=\sum\limits_{i=0}^{n}{{}^{n}{{C}_{i}}{{x}^{i}}{{y}^{n-i}}}\] and replace \[x\] by \[1\] and \[y\] by \[\dfrac{1}{2}a\]. Also, keep in mind to write \[{{z}^{-n}}=\dfrac{1}{{{z}^{n}}}\] as we can’t use binomial expansion for negative value of \[n\].

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 \[{{\left( x+y \right)}^{n}}=\sum\limits_{i=0}^{n}{{}^{n}{{C}_{i}}{{x}^{i}}{{y}^{n-i}}}\].
We will begin by writing \[{{\left( 1+\dfrac{1}{2}a \right)}^{-4}}\]as \[{{\left( 1+\dfrac{1}{2}a \right)}^{-4}}=\dfrac{1}{{{\left( 1+\dfrac{1}{2}a \right)}^{4}}}.....\left( 1 \right)\].
Substituting \[x=1,y=\dfrac{1}{2}a,n=4\] in the formula of binomial expansion \[{{\left( x+y \right)}^{n}}=\sum\limits_{i=0}^{n}{{}^{n}{{C}_{i}}{{x}^{i}}{{y}^{n-i}}}\], we have \[{{\left( 1+\dfrac{1}{2}a \right)}^{4}}={}^{4}{{C}_{0}}{{\left( 1 \right)}^{0}}{{\left( \dfrac{1}{2}a \right)}^{4}}+{}^{4}{{C}_{1}}{{\left( 1 \right)}^{1}}{{\left( \dfrac{1}{2}a \right)}^{3}}+{}^{4}{{C}_{2}}{{\left( 1 \right)}^{2}}{{\left( \dfrac{1}{2}a \right)}^{2}}+{}^{4}{{C}_{3}}{{\left( 1 \right)}^{3}}{{\left( \dfrac{1}{2}a \right)}^{1}}+{}^{4}{{C}_{4}}{{\left( 1 \right)}^{4}}{{\left( \dfrac{1}{2}a \right)}^{0}}.....\left( 2 \right)\]
We will now evaluate the values of \[{}^{4}{{C}_{i}}\forall i\in \{1,2,3,4\}\].
We know that \[{}^{x}{{C}_{y}}=\dfrac{x!}{y!\left( x-y \right)!}\]. We should also know that \[0!=1\].
Thus, we have \[{}^{4}{{C}_{0}}=\dfrac{4!}{0!\left( 4-0 \right)!}=\dfrac{4!}{4!}=1,{}^{4}{{C}_{1}}=\dfrac{4!}{1!\left( 4-1 \right)!}=\dfrac{4!}{3!}=\dfrac{4\times 3!}{3!}=4,{}^{4}{{C}_{2}}=\dfrac{4!}{2!2!}=\dfrac{4\times 3\times 2}{2\times 2}=6.....\left( 3 \right)\].
We know that \[{}^{x}{{C}_{y}}={}^{x}{{C}_{x-y}}\]. Thus, we have \[{}^{4}{{C}_{3}}={}^{4}{{C}_{4-3}}={}^{4}{{C}_{1}}=4,{}^{4}{{C}_{4}}={}^{4}{{C}_{4-4}}={}^{4}{{C}_{0}}=1.....\left( 4 \right)\].
Substituting the values from equation \[\left( 3 \right),\left( 4 \right)\] in equation \[\left( 2 \right)\], we get \[{{\left( 1+\dfrac{1}{2}a \right)}^{4}}=1{{\left( \dfrac{1}{2}a \right)}^{4}}+4{{\left( \dfrac{1}{2}a \right)}^{3}}+6{{\left( \dfrac{1}{2}a \right)}^{2}}+4{{\left( \dfrac{1}{2}a \right)}^{1}}+1{{\left( \dfrac{1}{2}a \right)}^{0}}\].
Simplifying the above equation, we have \[{{\left( 1+\dfrac{1}{2}a \right)}^{4}}=\dfrac{1}{16}{{a}^{4}}+\dfrac{1}{2}{{a}^{3}}+\dfrac{3}{2}{{a}^{2}}+2a+1.....\left( 5 \right)\].
Substituting the value of equation \[\left( 5 \right)\] in equation \[\left( 1 \right)\], we have \[{{\left( 1+\dfrac{1}{2}a \right)}^{-4}}=\dfrac{1}{{{\left( 1+\dfrac{1}{2}a \right)}^{4}}}=\dfrac{1}{\dfrac{1}{16}{{a}^{4}}+\dfrac{1}{2}{{a}^{3}}+\dfrac{3}{2}{{a}^{2}}+2a+1}\].
Hence, we get the expansion of \[{{\left( 1+\dfrac{1}{2}a \right)}^{-4}}\] upto \[4\] terms as \[\dfrac{1}{\dfrac{1}{16}{{a}^{4}}+\dfrac{1}{2}{{a}^{3}}+\dfrac{3}{2}{{a}^{2}}+2a+1}\].

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 \[n\].