Question

How do you expand ${\left( {d - \dfrac{1}{d}} \right)^5}?$

Answer 1

As we know that binomial theorem states that for any positive integer $n$, the nth power of the sum of the any two numbers $a$ and $b$ can be expressed as the sum of $(n + 1)$ terms of the form. The sum of the exponents of $a$ and $b$ or any given variable is always $n$. Here we have given an equation and we will expand it with the help of binomial theorem.

Complete step by step solution:
As per the given question we have an equation ${\left( {d - \dfrac{1}{d}} \right)^5}$ and we have to expand it. The binomial theorem states that ${(a + b)^n} = \sum\limits_{k = 0}^n {^n{C_k}} \cdot ({a^{n - k}}{b^k})$. Here $n$ is the exponential power and by using the combination and substituting the values we get: $\sum\limits_{k = 0}^5 {\dfrac{{5!}}{{(5 - k)!k!}}} \times {(d)^{5 - k}} \times {\left( { - \dfrac{1}{d}} \right)^k}$, By expanding this we get: $\dfrac{{5!}}{{(5 - 0)!0!}}{(d)^{5 - 0}} \times {(\dfrac{{ - 1}}{d})^0} + \dfrac{{5!}}{{(5 - 1)!0!}}{(d)^{5 - 1}} \times \left( { - \dfrac{1}{d}} \right) + ...$, since it will continue till the $n$factor.
Now we simplify the exponents for each term of the expansions, $1 \cdot {(d)^5} \times {\left( {\dfrac{{ - 1}}{d}} \right)^0} + 5 \times {(d)^4} \times \left( { - \dfrac{1}{d}} \right) + ... + 1 \times {(d)^0} \times {\left( { - \dfrac{1}{d}} \right)^5}$. It gives the polynomial equation i.e. ${d^5} - 5{d^3} + 10d - \dfrac{{10}}{d} - \dfrac{5}{{{d^3}}} - \dfrac{1}{{{d^5}}}$.

Hence the expanded form is ${d^5} - 5{d^3} + 10d - \dfrac{{10}}{d} - \dfrac{5}{{{d^3}}} - \dfrac{1}{{{d^5}}}$.

Note: We should always be careful while solving this kind of questions and we should know about the binomial theorem before solving this. We should note that each of the different groups or sections that can be formed by taking the same or all of the numbers irrespective of their arrangement is called the combination. We can understand this by example i.e. the different combinations formed by the letters $p,q,r$ is three in total. We should know that $pq$ and $qr$ are the arrangements which are called permutations but they give the same combination.