The greatest term in the expansion of $(1+x)^{15}$, when $x=\frac{1}{2}$ is
The greatest term in the expansion of $(1+x)^{15}$, when $x=\frac{1}{2}$ is
- $\frac{1}{32}{ }^{15} \mathrm{C}_5$
- $\frac{1}{64}{ }^{15} \mathrm{C}_6$
- $\frac{1}{32}{ }^{15} \mathrm{C}_6$
- $\frac{1}{64}{ }^{15} \mathrm{C}_5$
Solution
Let $T_{r+1}$ and $T_r$ denotes the $(\mathrm{r}+1)$ th and $r$ th term in the expansion of $(1+x)^{15}$
$
\therefore \quad T_{r+1}={ }^n C_r x^r \text { and } T_r={ }^n C_{r-1} x^{r-1} \text {, where } n=15
$
For greatest term in the expansion
$
\begin{aligned}
& T_{r+1} \geq T_r \Rightarrow{ }^n C_r x^r \geq{ }^n C_{r-1} x^{r-1} \\
& \Rightarrow \frac{n ! \cdot x^r}{r^{\prime} !(n-r) !} \geq \frac{n ! \cdot x^{r-1}}{(r-1) !(n-r+1) !}
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow \quad \frac{x}{r \cdot(r-1) !(\mathrm{n}-\mathrm{r}) !} \geq \frac{1}{(r-1) ! \cdot(n-r+1) \cdot(n-r) !} \\
& \Rightarrow \quad \frac{x}{r} \geq \frac{1}{(n-r+1)} \\
& \Rightarrow \frac{1}{r}\left(\frac{1}{2}\right) \geq \frac{1}{(16-r)} \quad\left(\because n=15 \text { and } x=\frac{1}{2}\right) \\
& \Rightarrow 16-r \geq 2 r \\
& \Rightarrow 3 r \leq 16 \Rightarrow r \leq \frac{16}{3}
\end{aligned}
$
Hence $r=5$ will be the greatest term
$
\therefore \quad T_{r+1}=T_{5+1}={ }^{15} C_5 x^5={ }^{15} C_5\left(\frac{1}{2}\right)^5=\frac{1}{32}{ }^{15} C_5
$
Asked in: AP EAMCET 2023 (19 May Shift 1)
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