Suppose 2-p, p, 2-α, α are coefficient of four consecutive terms in expansion of (1+x)^n . T

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7.Binomial Theorem

hard

Suppose $2-p, p, 2-\alpha, \alpha$ are the coefficient of four consecutive terms in the expansion of $(1+x)^n$. Then the value of $p^2-\alpha^2+6 \alpha+2 p$ equals

$4$

$10$

$8$

$6$

(JEE MAIN-2024)

Solution

$2-p, p, 2-\alpha, \alpha$

Binomial coefficients are

$\begin{array}{ll} & { }^n C_r,{ }^n C_{r+1},{ }^n C_{r+2},{ }^n C_{r+3} \text { respectively } \\ \Rightarrow & { }^n C_r+{ }^n C_{r+1}=2 \\ \Rightarrow & { }^{n+1} C_{r+1}=2 \quad \ldots \ldots .(1) \\ & \quad \text { Also, } \quad{ }^n C_{r+2}+{ }^n C_{r+3}=2 \\ \Rightarrow & { }^{n+1} C_{r+3}=2 \quad \ldots \ldots .(2) \\ & \quad \text { From }(1) \text { and (2) } \\ & { }^{n+1} C_{r+1}={ }^{n+1} C_{r+3}\end{array}$

$\begin{aligned} \Rightarrow \quad & 2 \mathrm{r}+4=\mathrm{n}+1 \\ & \mathrm{n}=2 \mathrm{r}+3 \\ & { }^{2 \mathrm{r}+4} \mathrm{C}_{\mathrm{r}+1}=2\end{aligned}$
Data Inconsistent

Standard 11

Mathematics

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hard

(JEE MAIN-2019)

If the term independent of $x$ in the exapansion of $\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{9}$ is $k,$ then $18 k$ is equal to

medium

(JEE MAIN-2020)

The coefficients of three consecutive terms in the expansion of $(1+a)^{n}$ are in the ratio $1: 7: 42 .$ Find $n$

hard

The term independent of $x$ in the expansion of $\left( {\frac{1}{{60}} – \frac{{{x^8}}}{{81}}} \right).{\left( {2{x^2} – \frac{3}{{{x^2}}}} \right)^6}$ is equal to

hard

(JEE MAIN-2019)

If in the expansion of ${(1 + x)^m}{(1 – x)^n}$, the coefficient of $x$ and ${x^2}$ are $3$ and $-6$ respectively, then m is

hard

(IIT-1999)

Suppose $2-p, p, 2-\alpha, \alpha$ are the coefficient of four consecutive terms in the expansion of $(1+x)^n$. Then the value of $p^2-\alpha^2+6 \alpha+2 p$ equals

Solution

Similar Questions

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