Question

The largest term is the expansion of $(3+2x{)}^{50}$ where $x=1/5$ is :

• A. $5^{th}$
• B. $6^{th}$
• C. $8^{th}$
• D. $9^{th}$

Answer 1

The correct option is B $6^{th}$

greatest term $(3+2x{)}^{50}$ $x=\frac{1}{5}$

$(3+2x{)}^{50}=(3+\frac{2}{5}{)}^{50}$

$=3^{50}(1+\frac{2}{15}{)}^{50}$

$(1+x)-\frac{Tr+1}{Tr}=\frac{n-r+1}{r}\times x$

For greatest

$(r+1{)}^{th}$ test is greatest

$tan\frac{Tr+1}{Tr}⩾1$

$In(1+\frac{2}{15}{)}^{50},\frac{Tr+1}{Tr}=\frac{50-r+1}{r}\times \frac{2}{15}$

$=\frac{(51-r)2}{15r}⩾1$

$(51-r)2⩾15r$

$102-2r⩾15r$

$102⩾17r$

$r⩽6$ (integer)

$T_r\&T_{r+1}$ both are greatest

$T_6\&T_7$

$3^{50}(1+\frac{2}{15}{)}^{50}$

$T_6=T_{5+1}=3^{50}{\times }^{50}{C}_5.(\frac{2}{15}{)}^5$

$T_9=T_{6+1}=3^{50}{\times }^{50}{C}_6\times (\frac{2}{15}{)}^6$

$\frac{T_7}{T_6}=\frac{{}^{50}{C}_5\times \frac{2}{15}}{{}^{50}{C}_5}$

$=\frac{50-6+1}{6}\times \frac{2}{15}$

$\frac{T_7}{T_6}=1$

$T_7=T_6$