माना 6 x की बढ़ती घातों में (3+6 x )^{ n } के | vedclass

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Question

$(3+6 x )^{ n }={ }^{ n } C _{0} 3^{ n }+{ }^{ n } C _{1} 3^{ n -1}(6 x )^{ I }+\ldots$

$T _{ r +1}{ }^{ n } C _{ r } 3^{ n - r } \cdot(6 x ) r $

Vedclass · Accepted Answer

$24$

Vedclass · Answer

$(3+6 x )^{ n }={ }^{ n } C _{0} 3^{ n }+{ }^{ n } C _{1} 3^{ n -1}(6 x )^{ I }+\ldots$ $T _{ r +1}{ }^{ n } C _{ r } 3^{ n - r } \cdot(6 x ) r $ $={ }^{ n } C _{ r } 3^{ n - r } \cdot 6^{ I } \cdot x ^{ r } 3^{ n - r } \cdot 3^{ r } \cdot 2^{ r } \cdot\left(\frac{3}{2} ight)^{ r }$ $={ }^{ n } C _{ r } 3^{ n } \cdot 3^{ r } \quad ext { [for } x =\frac{3}{2}]$ $T_{9}$ is greatest of $x =\frac{3}{2}$ So, $T _{9}> T _{10}$ and $T _{9}> T _{8}$ (concept of numerically greatest term) Here, $\frac{T_{9}}{T_{10}}>1$ and $\frac{T_{9}}{T_{8}}>1$ $\frac{{ }^{n} C_{8} 3^{n} \cdot 3^{8}}{{ }^{n} C_{9} 3^{n} \cdot 3^{9}}>1$ and $\frac{{ }^{n} C_{8} 3^{n} \cdot 3^{8}}{{ }^{n} C_{7} 3^{n} \cdot 3^{7}}>1$ and $\frac{{ }^{ n } C _{8}}{{ }^{ n } C _{7}}>\frac{1}{3}$ and $\frac{ n -7}{8}>\frac{1}{3}$ $\frac{29}{3}< n <11 \Rightarrow n =10= n _{0}$ So, in $(3+6 x )^{ n }$ for $n = n _{0}=10$ i.e., in $(3+6 x )^{10}$, here $T _{ r +1}={ }^{10} C _{ r } 3^{10- r } 6^{ T } x ^{ r }$ $T _{7}={ }^{10} C _{6} 3^{4} \cdot 6^{6} \cdot x ^{6}=210 \cdot 3^{10} \cdot 2^{6} x ^{6}$ $T _{4}={ }^{10} C _{3} 3^{7} 6^{3} x ^{3}=120.3^{10} \cdot 2^{3} x ^{3}$ Ratio of coefficient of $x ^{6}$ and coefficient of $x ^{3}= k$ $ herefore k =\frac{210 \cdot 3^{10} 2^{6}}{120 \cdot 3^{10} \cdot 2^{3}}=\frac{7}{4} imes 2^{3}=14$ So, $k + n _{0}=14+10=24$

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