If the coefficients of x^9, x^{10} and x^{11} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given that the coefficients of $x^9, x^{10}$ and $x^{11}$ in the expansion of $(1+x)^n$ are in $A$.$P$.This means ${}^nC_9, {}^nC_{10}$ and ${}^nC

Vedclass · Accepted Answer

$-398$

Vedclass · Answer

(C) Given that the coefficients of $x^9, x^{10}$ and $x^{11}$ in the expansion of $(1+x)^n$ are in $A$.$P$.This means ${}^nC_9, {}^nC_{10}$ and ${}^nC_{11}$ are in $A$.$P$.Therefore,$2({}^nC_{10}) = {}^nC_9 + {}^nC_{11}$.Dividing by ${}^nC_{10}$,we get $2 = \frac{{}^nC_9}{{}^nC_{10}} + \frac{{}^nC_{11}}{{}^nC_{10}}$.Using the formula $\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n-r+1}{r}$,we have:$2 = \frac{10}{n-9} + \frac{n-10}{11}$.$2 = \frac{110 + (n-10)(n-9)}{11(n-9)}$.$22(n-9) = 110 + n^2 - 19n + 90$.$22n - 198 = n^2 - 19n + 200$.$n^2 - 41n = -198 - 200 = -398$.

If the coefficients of $x^9, x^{10}$ and $x^{11}$ in the expansion of $(1+x)^n$ are in arithmetic progression,then $n^2-41n$ is equal to

Similar Questions

If $C_r = { }^n C_r$,then find the sum $C_0 + C_4 + C_8 + C_{12} + \ldots$

If $^nC_0 + \frac{1}{2} ^nC_1 + \frac{1}{3} ^nC_2 + \dots + \frac{1}{n+1} ^nC_n = \frac{1023}{10}$, then $n =$ ?

If $(1 + x)^n = \sum\limits_{r = 0}^n {{C_r}{x^r}} $,then $\left( {1 + \frac{{{C_1}}}{{{C_0}}}} \right)\left( {1 + \frac{{{C_2}}}{{{C_1}}}} \right)....\left( {1 + \frac{{{C_n}}}{{{C_{n - 1}}}}} \right) = $

If $n \geq 2$ is a positive integer,then the sum of the series ${ }^{n+1} C_{2}+2\left({ }^{2} C_{2}+{ }^{3} C_{2}+{ }^{4} C_{2}+\ldots+{ }^{n} C_{2}\right)$ is ...... .

The value of $\sum\limits_{r = 0}^{n - 1} {\frac{{^n{C_r}}}{{^n{C_r} + {\,^n}{C_{r + 1}}}}} $ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module