The sum of the binomial coefficient of ${\left[ {2x + \dfrac{1}{x}} \right]^n}$ is equal to 256. The constant term in the expansion is
A.1120
B.2110
C.1210
D.None of the above

Question

The sum of the binomial coefficient of ${\left[ {2x + \dfrac{1}{x}} \right]^n}$ is equal to 256. The constant term in the expansion isA.1120B.2110C.1210 D.None of the above

Vedantu Content Team · Accepted Answer

Hint: The sum of the binomial coefficient is ${2^n}$ and ${t_{r + 1}}$ is called a general term for all $r \in N$ and $0 \leqslant r \leqslant n$ . Using this formula, you can find any term of the expansion.Complete step-by-step answer:Given that, the expansion ${\left[ {2x + \dfrac{1}{x}} \right]^n}$ and the sum of the binomial coefficient = 256.We know that,The sum of the binomial coefficient $ = {2^n} $, where n is the number of terms$256 = {2^n}$We have $256 = {2^8}$${2^8} = {2^n}$$8 = n$Hence $n = 8$Now, comparing the given expansion ${\left[ {2x + \dfrac{1}{x}} \right]^n}$ with the expansion ${\left( {a + b} \right)^n}$ and then we have$a = 2x$ and $b = \dfrac{1}{x}$The formula for general term for the expansion ${\left( {a + b} \right)^n}$ is given by${t_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}.................(1),{\text{ for all }}r \in N{\text{ and }}0 \leqslant r \leqslant n$Here, $a = 2x,{\text{ }}b = \dfrac{1}{x}{\text{ and }}n = 8$Now put all the values in the equation (1), we get${t_{r + 1}} = {}^8{C_r}{(2x)^{8 - r}}{\left( {\dfrac{1}{x}} \right)^r}$The rearranging the terms by using the indices formulae, we get${t_{r + 1}} = {}^8{C_r}{(2)^{8 - r}}{\left( x \right)^{8 - r}}\dfrac{1}{{{x^r}}}$$\Rightarrow {t_{r + 1}} = {}^8{C_r}{(2)^{8 - r}}{\left( x \right)^{8 - r}}{(x)^{ - r}}$$\Rightarrow {t_{r + 1}} = {}^8{C_r}{(2)^{8 - r}}{\left( x \right)^{8 - r - r}}$$\Rightarrow {t_{r + 1}} = {}^8{C_r}{(2)^{8 - r}}{\left( x \right)^{8 - 2r}}...........(2)$To get the constant term that means the term independent of x, we must have${x^{8 - 2r}} = {x^0}$$\Rightarrow 8 - 2r = 0$$\Rightarrow 2r = 8$Dividing both sides by 2, we get$r = 4$The constant term in the equation (2) is given byConstant term $ = {}^8{C_r}{(2)^{8 - r}}.............(3)$Now put r = 4 in the equation (3), we getConstant term $ = {}^8{C_4}{(2)^{8 - 4}}$Constant term $ = {}^8{C_4}{(2)^4}$We have${2^4} = 16$Constant term $ = 16 {}^8{C_4}..............(4)$We know that the formula for combination, ${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$${}^8{C_4} = \dfrac{{8!}}{{4!(8 - 4)!}} = \dfrac{{8!}}{{4! \times 4!}} = \dfrac{{8 \times 7 \times 6 \times 5}}{{1 \times 2 \times 3 \times 4}} = \dfrac{{7 \times 2 \times 5}}{1} = 70$Now put the value of ${}^8{C_2}$ in the equation (4), we getConstant term $ = 16 \times 70$Constant term in the given expansion $ = 1120$Hence the correct option of the given question is option (a).Note: The possibility for the mistake is that you might get confused for finding the constant term in the given expansion. The constant term in the given expansion that means to get the term independent of x.