Question

Find the term independent of ‘x’ in the expansion of ${\left( {4{x^3} + \dfrac{1}{2}x} \right)^8}$.

Answer 1

Hint: The given question requires us to find the term independent of x in the binomial expansion ${\left( {4{x^3} + \dfrac{1}{2}x} \right)^8}$ . Binomial theorem helps us to expand the powers of binomial expressions easily and can be used to solve the given problem. We must know the formulae of combinations and factorials for solving the given question using the binomial theorem. To solve the problem, we find the general term of the binomial expansion and then equate the power of x as zero.

Complete step by step answer:
We are given the binomial expression ${\left( {4{x^3} + \dfrac{1}{2}x} \right)^8}$.
So, using the binomial theorem, the binomial expansion of \[{\left( {x + y} \right)^n}\] is
$\sum\nolimits_{r = 0}^n {\left( {^n{C_r}} \right){{\left( x \right)}^{n - r}}{{\left( y \right)}^r}} $
So, the binomial expansion of ${\left( {4{x^3} + \dfrac{1}{2}x} \right)^8}$ is \[\sum\nolimits_{r = 0}^8 {\left( {^8{C_r}} \right){{\left( {4{x^3}} \right)}^{8 - r}}{{\left( {\dfrac{1}{2}x} \right)}^r}} \] .
Now, we know that the general term of the binomial expansion of the expression ${\left( {4{x^3} + \dfrac{1}{2}x} \right)^8}$ is \[\left( {^8{C_r}} \right){\left( {4{x^3}} \right)^{8 - r}}{\left( {\dfrac{1}{2}x} \right)^r}\].

Simplifying the expression and separating the constants, we get,
\[ \Rightarrow \left( {^8{C_r}} \right){\left( 4 \right)^{8 - r}}{\left( {{x^3}} \right)^{8 - r}}{\left( {\dfrac{1}{2}} \right)^r}{x^r}\]
Now, we simplify this expression and compute the power of x,
\[ \Rightarrow \left( {^8{C_r}} \right){\left( 4 \right)^{8 - r}}{\left( {\dfrac{1}{2}} \right)^r}\left( {{x^{24 - 3r + r}}} \right)\]
Simplifying the expression further, we get,
\[ \Rightarrow \left( {^8{C_r}} \right){\left( 4 \right)^{8 - r}}{\left( {\dfrac{1}{2}} \right)^r}\left( {{x^{24 - 2r}}} \right)\]
Now, we equate the power of x to zero so as to find the term independent of x.
\[24 - 2r = 0\]
Now, shifting the terms in the equation to find the value of r, we get,
\[ \Rightarrow 2r = 24\]

Dividing both sides by two, we get,
\[ \Rightarrow r = \dfrac{{24}}{2}\]
Cancelling common factors in numerator and denominator, we get,
\[ \Rightarrow r = 12\]
Now, we notice that the value of r is even greater than the whole power of the original term ${\left( {4{x^3} + \dfrac{1}{2}x} \right)^8}$.
We know the expansion of the given binomial expression as \[\sum\nolimits_{r = 0}^8 {\left( {^8{C_r}} \right){{\left( {4{x^3}} \right)}^{8 - r}}{{\left( {\dfrac{1}{2}x} \right)}^r}} \]. But the value of r in the combination formula $^n{C_r}$ cannot be greater than n. So, n should be less than or equal to $8$.Hence, we cannot put \[r = 12\] in the combination formula.

Hence, the term independent of $x$ in the expansion of ${\left( {4{x^3} + \dfrac{1}{2}x} \right)^8}$ does not exist.

Note: If the value of r would have been in the specified limits such that we could put the value in the combination formula $^n{C_r}$, then we would have found the term independent of x simply by substituting the value of r in the general term formula of a binomial expansion as $\left( {^n{C_r}} \right){\left( x \right)^{n - r}}{\left( y \right)^r}$.