The greatest term in the expansion of (1+x)^{ | vedclass

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(A) Let $T_{r+1}$ and $T_r$ denote the $(r+1)$-th and $r$-th terms in the expansion of $(1+x)^{15}$.Given $n=15$ and $x=\frac{1}{2}$.For the greatest

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$\frac{1}{32}{ }^{15} C_5$

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(A) Let $T_{r+1}$ and $T_r$ denote the $(r+1)$-th and $r$-th terms in the expansion of $(1+x)^{15}$.Given $n=15$ and $x=\frac{1}{2}$.For the greatest term,we have the condition $T_{r+1} \geq T_r$.This implies $\frac{T_{r+1}}{T_r} \geq 1$.Using the formula for the general term,$\frac{T_{r+1}}{T_r} = \frac{{}^{n}C_r x^r}{{}^{n}C_{r-1} x^{r-1}} = \frac{n-r+1}{r} \cdot x$.Substituting the values: $\frac{15-r+1}{r} \cdot \frac{1}{2} \geq 1$.$\frac{16-r}{2r} \geq 1$.$16-r \geq 2r$.$3r \leq 16 \Rightarrow r \leq \frac{16}{3} \approx 5.33$.Since $r$ must be an integer,the greatest term occurs at $r=5$,which is $T_{5+1} = T_6$.$T_6 = {}^{15}C_5 \cdot x^5 = {}^{15}C_5 \cdot (\frac{1}{2})^5 = \frac{1}{32} {}^{15}C_5$.

The greatest term in the expansion of $(1+x)^{15}$,when $x=\frac{1}{2}$ is

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