If the coefficients of r-th and (r+1)-th term | vedclass

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

Question

(D) The $r$-th term in the expansion of $(1+x)^{24}$ is $T_r = {}^{24}C_{r-1} x^{r-1}$,so its coefficient is ${}^{24}C_{r-1}$.The $(r+1)$-th term is $

Vedclass · Accepted Answer

$x^2-14x+24=0$

Vedclass · Answer

(D) The $r$-th term in the expansion of $(1+x)^{24}$ is $T_r = {}^{24}C_{r-1} x^{r-1}$,so its coefficient is ${}^{24}C_{r-1}$.The $(r+1)$-th term is $T_{r+1} = {}^{24}C_r x^r$,so its coefficient is ${}^{24}C_r$.Given the ratio of coefficients is $12:13$,we have $\frac{{}^{24}C_{r-1}}{{}^{24}C_r} = \frac{12}{13}$.Using the formula $\frac{{}^nC_{k-1}}{{}^nC_k} = \frac{k}{n-k+1}$,we get $\frac{r}{24-r+1} = \frac{12}{13}$.$\frac{r}{25-r} = \frac{12}{13} \implies 13r = 300 - 12r \implies 25r = 300 \implies r = 12$.Now,checking the quadratic equations for $r=12$:For $x^2-14x+24=0$,substituting $x=12$ gives $144 - 168 + 24 = 0$,which is true.Thus,$r=12$ is a root of the equation $x^2-14x+24=0$.

If the coefficients of $r$-th and $(r+1)$-th terms in the expansion of $(1+x)^{24}$ are in the ratio $12:13$,then $r$ is the root of the quadratic equation

Similar Questions

Find the term independent of $x$ in the expansion of $\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{6}$.

The coefficient of $x^{24}$ in the expansion of $(1+x^2)^{12}(1+x^{12})(1+x^{24})$ is

The value of $x$ in the expression $(x + x^{\log_{10} x})^5$,if the third term in the expansion is $1,000,000$.

The coefficient of $x^5$ in the expansion of $(1+x)^{21}+(1+x)^{22}+\ldots+(1+x)^{30}$ is

The coefficient of ${x^{-9}}$ in the expansion of ${\left( \frac{x^2}{2} - \frac{2}{x} \right)^9}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module