For mathrm{r}=0,1, ldots, 10 , let mathrm{A}_{mathrm{r}}, mathrm{B}_{mathrm{r}} and mathrm{C}

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7.Binomial Theorem

hard

For $\mathrm{r}=0,1, \ldots, 10$, let $\mathrm{A}_{\mathrm{r}}, \mathrm{B}_{\mathrm{r}}$ and $\mathrm{C}_{\mathrm{r}}$ denote, respectively, the coefficient of $\mathrm{x}^{\mathrm{r}}$ in the expansions of $(1+\mathrm{x})^{10}$, $(1+\mathrm{x})^{20}$ and $(1+\mathrm{x})^{30}$. Then $\sum_{r=1}^{10} A_r\left(B_{10} B_r-C_{10} A_r\right)$ is equal to

$\mathrm{B}_{10}-\mathrm{C}_{10}$

$A_{10}\left(B_{10}^2-C_{10} A_{10}\right)$

$0$

$\mathrm{C}_{10}-\mathrm{B}_{10}$

(IIT-2010)

Solution

$ \text { Let } \mathrm{y}=\sum_{\mathrm{r}=1}^{10} \mathrm{~A}_{\mathrm{r}}\left(\mathrm{B}_{10} \mathrm{~B}_{\mathrm{r}}-\mathrm{C}_{10} \mathrm{~A}_{\mathrm{r}}\right) $

$ \sum_{\mathrm{r}=1}^{10} \mathrm{~A}_{\mathrm{r}} \mathrm{B}_{\mathrm{r}}=\text { coefficient of } \mathrm{x}^{20} \text { in }\left((1+\mathrm{x})^{10}(\mathrm{x}+1)^{20}\right)-1 $

$ =C_{20}-1=\mathrm{C}_{10}-1 \text { and } \sum_{\mathrm{r}=1}^{10}\left(\mathrm{~A}_{\mathrm{r}}\right)^2=\text { coefficient of } \mathrm{x}^{10} \text { in }\left((1+\mathrm{x})^{10}(\mathrm{x}+1)^{10}\right)-1=\mathrm{B}_{10}-1 $

$ \Rightarrow \mathrm{y}=\mathrm{B}_{10}\left(\mathrm{C}_{10}-1\right)-\mathrm{C}_{10}\left(\mathrm{~B}_{10}-1\right)=\mathrm{C}_{10}-\mathrm{B}_{10} .$

Standard 11

Mathematics

The term independent of $x$ in the expansion of ${\left( {2x – \frac{3}{x}} \right)^6}$ is

easy

Let the sum of the coefficients of the first three terms in the expansion of $\left(x-\frac{3}{x^2}\right)^n, x \neq 0, n \in N$, be $376$. Then the coefficient of $x^4$ is $……$

hard

(JEE MAIN-2023)

In the binomial $(2^{1/3} + 3^{-1/3})^n$, if the ratio of the seventh term from the beginning of the expansion to the seventh term from its end is $1/6$ , then $n =$

normal

The smallest natural number $n,$ such that the coefficient of $x$ in the expansion of ${\left( {{x^2}\, + \,\frac{1}{{{x^3}}}} \right)^n}$ is $^n{C_{23}}$ is

hard

(JEE MAIN-2019)

If the coefficient of the middle term in the expansion of ${(1 + x)^{2n + 2}}$ is $p$ and the coefficients of middle terms in the expansion of ${(1 + x)^{2n + 1}}$ are $q$ and $r$, then

medium

Solution

Similar Questions

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