Let α=Σ_{k=0}^nleft(frac{left({ }^n Cₖright)^2}{k+1}right) and β=Σ

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7.Binomial Theorem

normal

Let $\alpha=\sum_{k=0}^n\left(\frac{\left({ }^n C_k\right)^2}{k+1}\right)$ and $\beta=\sum_{k=0}^{n-1}\left(\frac{{ }^n C_k{ }^n C_{k+1}}{k+2}\right)$. If $5 \alpha=6 \beta$, then $n$ equals

$6$

$7$

$9$

$10$

(JEE MAIN-2024)

Solution

$\sum_{k=0}^n \frac{{ }^n C_k \cdot{ }^n C_k}{k+1} \cdot \frac{n+1}{n+1} $

$ =\frac{1}{n+1} \sum_{k=0}^n{ }^{n+1} C_{k+1} \cdot{ }^n C_{n-k} $

$ =\frac{1}{n+1} \cdot{ }^{2 n+1} C_{n+1} $

$ =\sum_{k=0}^{n-1}{ }^n C_k \cdot \frac{{ }^n C_{k+1}}{k+2} \frac{n+1}{n+1} $

$ \frac{1}{n+1} \sum_{k=0}^{n-1} C_{n-k} \cdot{ }^{n+1} C_{k+2} $

$ =\frac{1}{n+1} \cdot{ }^{2 n+1} C_{n+2}$

$\alpha=\frac{1}{n+1} \cdot{ }^{2 n+1} C_{n+1}$

$ \beta=\sum_{k=0}^{n-1} C_k \cdot \frac{{ }^n C_{k+1}}{k+2} \frac{n+1}{n+1} $

$ \frac{1}{n+1} \sum_{k=0}^{n-1} C_{n-k} \cdot{ }^{n+1} C_{k+2}$

$ =\frac{1}{n+1} \cdot{ }^{2 n+1} C_{n+2} $

$ \frac{\beta}{\alpha}=\frac{{ }^{2 n+1} C_{n+2}}{{ }^{2 n+1} C_{n+1}}=\frac{2 n+1-(n+2)+1}{n+2} $

$ \frac{\beta}{\alpha}=\frac{n}{n+2}=\frac{5}{6} $

$n=10$

Standard 11

Mathematics

Suppose $\sum \limits_{ r =0}^{2023} r ^{20023} C _{ r }=2023 \times \alpha \times 2^{2022}$. Then the value of $\alpha$ is $…………$

hard

(JEE MAIN-2023)

If $1+\left(2+{ }^{49} C _{1}+{ }^{49} C _{2}+\ldots .+{ }^{49} C _{49}\right)\left({ }^{50} C _{2}+{ }^{50} C _{4}+\right.$ $\ldots . .+{ }^{50} C _{ so }$ ) is equal to $2^{ n } . m$, where $m$ is odd, then $n$ $+m$ is equal to.

hard

(JEE MAIN-2022)

If ${(1 + x)^{15}} = {C_0} + {C_1}x + {C_2}{x^2} + …… + {C_{15}}{x^{15}},$ then ${C_2} + 2{C_3} + 3{C_4} + …. + 14{C_{15}} = $

hard

(IIT-1966)

Let ${ }^{n} C_{r}$ denote the binomial coefficient of $x^{r}$ in the expansion of $(1+ x )^{ n }.$

If $\sum_{ k =0}^{10}\left(2^{2}+3 k \right){ }^{ n } C _{ k }=\alpha .3^{10}+\beta \cdot 2^{10}, \alpha, \beta \in R$ then $\alpha+\beta$ is equal to ……. .

hard

(JEE MAIN-2021)

The coefficient of $x^8$ in the expansion of $(x-1) (x- 2) (x-3)……………(x-10)$ is :

normal

Let $\alpha=\sum_{k=0}^n\left(\frac{\left({ }^n C_k\right)^2}{k+1}\right)$ and $\beta=\sum_{k=0}^{n-1}\left(\frac{{ }^n C_k{ }^n C_{k+1}}{k+2}\right)$. If $5 \alpha=6 \beta$, then $n$ equals

Solution

Similar Questions

Suppose $\sum \limits_{ r =0}^{2023} r ^{20023} C _{ r }=2023 \times \alpha \times 2^{2022}$. Then the value of $\alpha$ is $…………$

If $1+\left(2+{ }^{49} C _{1}+{ }^{49} C _{2}+\ldots .+{ }^{49} C _{49}\right)\left({ }^{50} C _{2}+{ }^{50} C _{4}+\right.$ $\ldots . .+{ }^{50} C _{ so }$ ) is equal to $2^{ n } . m$, where $m$ is odd, then $n$ $+m$ is equal to.

If ${(1 + x)^{15}} = {C_0} + {C_1}x + {C_2}{x^2} + …… + {C_{15}}{x^{15}},$ then ${C_2} + 2{C_3} + 3{C_4} + …. + 14{C_{15}} = $

Let ${ }^{n} C_{r}$ denote the binomial coefficient of $x^{r}$ in the expansion of $(1+ x )^{ n }.$ If $\sum_{ k =0}^{10}\left(2^{2}+3 k \right){ }^{ n } C _{ k }=\alpha .3^{10}+\beta \cdot 2^{10}, \alpha, \beta \in R$ then $\alpha+\beta$ is equal to ……. .

The coefficient of $x^8$ in the expansion of $(x-1) (x- 2) (x-3)……………(x-10)$ is :

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Let ${ }^{n} C_{r}$ denote the binomial coefficient of $x^{r}$ in the expansion of $(1+ x )^{ n }.$

If $\sum_{ k =0}^{10}\left(2^{2}+3 k \right){ }^{ n } C _{ k }=\alpha .3^{10}+\beta \cdot 2^{10}, \alpha, \beta \in R$ then $\alpha+\beta$ is equal to ……. .