Question

The seventh term in the expansion of
${(4x-\frac{1}{2\sqrt{x}})}^{13}$ is

• A. ${}^{13}{C}_7{.2}^5.x^4$
• B. ${}^{13}{C}_6{.2}^8.x^3$
• C. ${}^{13}{C}_6{.2}^8.x^4$
• D. ${}^{13}{C}_6{.2}^6.x^5$

Answer 1

Answer: (C)

${}^{13}{C}_6{.2}^8.x^4$

$(x-a{)}^n={}^n{C}_0{x}^n-{}^n{C}_1{x}^{n-1}a+{}^n{C}_2{x}^{n-2}{a}^2-.....$

$(r+1)$th term in the expansion of ${(x-a)}^n$ is given by

$T_{r+1}=(-1{)}^r{}^n{C}_r{x}^{n-r}{a}^r$

Using above formula for ${(4x-\frac{1}{2\sqrt{x}})}^{13}$, seventh term in the expansion is

$T_7=T_{6+1}=\left(-1{)}^6{}^{13}{C}_6\right(4x{)}^{13-6}{\left(\frac{1}{2\sqrt{x}}\right)}^6$

$={}^{13}{C}_6(4x{)}^7\left(\frac{1}{2^6(\sqrt{x}{)}^6}\right)$

$={}^{13}{C}_6\cdot 2^{14}\cdot x^7\left(\frac{1}{2^6{x}^3}\right)$

$={}^{13}{C}_6\cdot 2^8\cdot x^4$