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# Expand to 4 terms the following expressions: ${{\left( 1+\dfrac{1}{2}a \right)}^{-4}}$  Verified
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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 $n$.