Ratio of coefficient of x^{15} to term independent of x in expansion of {left( {{x^2} + frac{2

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7.Binomial Theorem

hard

The ratio of the coefficient of $x^{15}$ to the term independent of $x$ in the expansion of ${\left( {{x^2} + \frac{2}{x}} \right)^{15}}$ is

$7: 16$

$7:64$

$1: 4$

$1: 32$

(JEE MAIN-2013)

Solution

$T_{r+1}=^{15} C_{r}\left(x^{2}\right)^{15-r} \cdot\left(2 x^{-1}\right)^r$

$=15 \mathrm{C}_{r} \times(2)^{r} \times x^{30-3r}$

For independent term, $30-3 r=0$

$\Rightarrow r=10$

Hence the term independent of $x$.

$\mathrm{T}_{11}=^{15} \mathrm{C}_{10} \times(2)^{10}$

For term involve $x^{15}, 30-3 r=15$

$ \Rightarrow r=5$

Hence coefficient of $x^{15}=^{15} C_{5} \times(2)^{5}$

Required ratio

$ = \frac{{{\text{15}}{{\text{C}}_5} \times {{(2)}^5}}}{{15{C_{10}} \times {{(2)}^{10}}}} = \frac{{\frac{{15!}}{{10!5!}}}}{{\frac{{15!}}{{5!10!}} \times {{(2)}^5}}}$

$ = 1:32$

Standard 11

Mathematics

If the fourth term in the binomial expansion of $\left(\sqrt{\frac{1}{x^{1+\log _{10} x}}}+x^{\frac{1}{12}}\right)^{6}$ is equal to $200$, and $x > 1$, then the value of $x$ is

hard

(JEE MAIN-2019)

Let $S=\{a+b \sqrt{2}: a, b \in Z \}, T_1=\left\{(-1+\sqrt{2})^n: n \in N \right\}$ and $T_2=\left\{(1+\sqrt{2})^n: n \in N \right\}$. Then which of the following statements is (are) $TRUE$?

$(A)$ $Z \cup T_1 \cup T_2 \subset S$

$(B)$ $T_1 \cap\left(0, \frac{1}{2024}\right)=\phi$, where $\phi$ denotes the empty set

$(C)$ $T_2 \cap(2024, \infty) \neq \phi$

$(D)$ For any given $a, b \in Z , \cos (\pi(a+b \sqrt{2}))+i \sin (\pi(a+b \sqrt{2})) \in Z$ if and only if $b=0$, where $i=\sqrt{-1}$

normal

(IIT-2024)

Find the $r^{\text {th }}$ term from the end in the expansion of $(x+a)^{n}$

medium

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

hard

The ratio of coefficient of $x^2$ to coefficient of $x^{10}$ in the expansion of ${\left( {{x^5} + {{4.3}^{ – {{\log }_{\sqrt 3 }}\sqrt {{x^3}} }}} \right)^{10}}$ is

normal

The ratio of the coefficient of $x^{15}$ to the term independent of $x$ in the expansion of ${\left( {{x^2} + \frac{2}{x}} \right)^{15}}$ is

Solution

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