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

$(1 + x) (1 + x + x^2) (1 + x + x^2 + x^3) …… (1 + x + x^2 + …… + x^{100})$ when written in the ascending power of $x$ then the highest exponent of $x$ is ______ .

normal

$\sum_{\substack{i, j=0 \\ i \neq j}}^{n}{ }^{n} C_{i}{ }^{n} C_{j}$ is equal to

normal

(JEE MAIN-2022)

If the expansion in powers of $x$ of the function $\frac{1}{{\left( {1 – ax} \right)\left( {1 – bx} \right)}}$ is ${a_0} + {a_1}x + {a_2}{x^2} + \;{a_3}{x^3} + \; \ldots……$ then ${a_n}$ is

hard

(AIEEE-2006)

If ${S_n} = \sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}}} $ and ${t_n} = \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} $, then $\frac{{{t_n}}}{{{S_n}}}$ is equal to

hard

(AIEEE-2004)

If ${\left( {1 + x} \right)^n} = {c_0} + {c_1}x + {c_2}{x^2} + {c_3}{x^3} + …… + {c_n}{x^n}$ , then the value of ${c_0} – 3{c_1} + 5{c_2} – …….. + {( – 1)^n}\,(2n + 1){c_n}$ 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

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