Question

If the \[{r^{th}}\] term is the middle term in the expansion of ${\left( {{x^2} - \dfrac{1}{{2x}}} \right)^{20}}$ then the ${\left( {r + 3} \right)^{th}}$ term is
A) ${}^{20}{C_{14}} - \dfrac{1}{{{2^{14}}}} \times x$
B) ${}^{20}{C_{12}} - \dfrac{1}{{{2^{12}}}} \times {x^2}$
C) $ - \dfrac{1}{{{2^{13}}}} \times {}^{20}{C_7} \times x$
D) None of these

Answer 1

Hint:
We can find the total number of terms in the expansion and then find the value of r by dividing the total number of terms. Then we can use the binomial expansion to find the ${\left( {r + 3} \right)^{th}}$ term. We can obtain the required answer after further simplification.

Complete step by step solution:
We know that the binomial expansion of ${\left( {a + b} \right)^n}$ has $n + 1$ terms.
So, the expansion of ${\left( {{x^2} - \dfrac{1}{{2x}}} \right)^{20}}$ will have $20 + 1 = 21$ terms.
As there are an odd number of terms, the middle most term is given by, \[{\dfrac{{n + 1}}{2}^{th}}\] term.
We are given that \[{r^{th}}\] term is the middle term. So, we can write,
 $ \Rightarrow r = \dfrac{{n + 1}}{2}$
On substituting the value of n, we get,
 $ \Rightarrow r = \dfrac{{21 + 1}}{2}$
On simplification we get,
 $ \Rightarrow r = \dfrac{{22}}{2}$
On division we get,
 $ \Rightarrow r = 11$
We need to find the ${\left( {r + 3} \right)^{th}}$ term. So, we need to find $\left( {11 + 3} \right) = {14^{th}}$ term
We know that ${\left( {r + 1} \right)^{th}}$ term of a binomial expansion of ${\left( {a + b} \right)^n}$ is given by,
 \[{t_{r + 1}} = {}^n{C_r} \times {a^{n - r}} \times {b^r}\]
So, the \[{14^{th}}\] term is given by,
 $ \Rightarrow {t_{13 + 1}} = {}^n{C_{13}} \times {a^{n - 13}} \times {b^{13}}$
We have $n = 20$, $a = {x^2}$ and $b = - \dfrac{1}{{2x}}$. On substituting these values, we get,
 $ \Rightarrow {t_{14}} = {}^{20}{C_{13}} \times {\left( {{x^2}} \right)^{20 - 13}} \times {\left( {\dfrac{{ - 1}}{{2x}}} \right)^{13}}$
On simplifying using the properties of exponents, we get,
 $ \Rightarrow {t_{14}} = {}^{20}{C_{13}} \times {\left( x \right)^{2 \times 7}} \times \dfrac{{{{\left( { - 1} \right)}^{13}}}}{{{2^{13}} \times {x^{13}}}}$
We know that ${}^n{C_r} = {}^n{C_{n - r}}$, so we get,
 $ \Rightarrow {t_{14}} = {}^{20}{C_{20 - 13}} \times {\left( x \right)^{14 - 13}} \times \dfrac{{ - 1}}{{{2^{ - 13}}}}$
On simplification, we get,
 $ \Rightarrow {t_{14}} = - {}^{20}{C_7} \times x \times \dfrac{1}{{{2^{ - 13}}}}$
So, we have,
 \[ \Rightarrow {t_{14}} = - \dfrac{1}{{{2^{ - 13}}}} \times {}^{20}{C_7} \times x\]
Therefore, the ${\left( {r + 3} \right)^{th}}$ term is $ - \dfrac{1}{{{2^{13}}}} \times {}^{20}{C_7} \times x$

So, the correct answer is option C.

Note:
We must note that equation of the ${\left( {r + 1} \right)^{th}}$ term of a binomial expansion of ${\left( {a + b} \right)^n}$ is given by, ${}^n{C_r} \times {a^{n - r}} \times {b^r}$ not the \[{r^{th}}\] term. The binomial expansion of ${\left( {a + b} \right)^n}$ is given by, \[{\left( {a + b} \right)^n} = {}^n{C_0} \times {a^n} \times {b^0} + {}^n{C_1} \times {a^{n - 1}} \times {b^1} + {}^n{C_2} \times {a^{n - 2}} \times {b^2} + ...... + {}^n{C_n}v \times {a^0} \times {b^n}\]. We need not to find the middle term. We just need to find the value of r by finding the position of the middle term in the expansion. We must take the power of -1 as positive for even powers and negative for odd powers.