Let for the 9^{text {th }} term in the binomi | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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, $k + n _{0}=14+10=24$

Similar Questions

The coefficient of ${x^5}$ in the expansion of ${({x^2} - x - 2)^5}$ is

If the greatest value of the term independent of $^{\prime}x^{\prime}$ in the expansion of $\left(x \sin \alpha+a \frac{\cos \alpha}{x}\right)^{10}$ is $\frac{10 !}{(5 !)^{2}}$, then the value of $' a^{\prime}$ is equal to:

If $p$ and $q$ be positive, then the coefficients of ${x^p}$ and ${x^q}$ in the expansion of ${(1 + x)^{p + q}}$will be

Let $m$ be the smallest positive integer such that the coefficient of $x^2$ in the expansion of $(1+x)^2+(1+x)^3+\cdots+(1+x)^{49}+(1+m x)^{50}$ is $(3 n+1)^{51} C_3$ for some positive integer $n$. Then the value of $n$ is