Question

The power of $x$ in the term with the greatest coefficient in the expansion of $\left(1+\frac{x}{2}\right)^{10}$ is:

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

The Correct Option is B

$\left(1+\frac{x}{2}\right)^{10}=1+{ }^{10} C_{1} \frac{1}{2} x$
$+{ }^{10} C_{2}\left(\frac{1}{2} x\right)^{2}+{ }^{10} C_{3}\left(\frac{1}{2} x\right)^{3}+{ }^{10} C_{4}\left(\frac{1}{2} x\right)^{4}+\ldots$
By inspection, ${ }^{10} C_{3} \cdot \frac{1}{8}$ is the highest coefficient.
So, power of the $x$ is 3 .

Answer 2

The efficient of \((r+1)^{th}\) term be the greatest coefficient.
\((\frac {T_{r+1}}{T_r})_{x=1}> 1\) and \((\frac {T_{r+2}}{T_{r+1}})_{x=1} < 1\)

\(\frac {^{10}C_r(\frac 12)^r}{^{10}C_{r-1}(\frac 12)^{r-1}}>1\) and \(\frac {^{10}C_{r+1}(\frac 12)^{r+1}}{^{10}C_{r}(\frac 12)^{r}}<1\)

\(\frac {11-r}{2r} >1\) and \(\frac {10-r}{2(r+1)}<1\)
\(8<3r<11\)
\(r=3\)
Now, the greatest coefficient,
\(T_{r+1}=^{10}C_r(\frac 12)^r\)
Put r=3,
\(T_{4}=^{10}C_3(\frac 12)^3 x^3\)
Then, the power of coefficient with greatest coefficient = 3

So, the correct option is (B): 3