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

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(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 $

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

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(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

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