If a_r is the coefficient of x^{10-r} in the | vedclass

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(B) Given $a_r$ is the coefficient of $x^{10-r}$ in $(1+x)^{10}$,we have $a_r = {}^{10}C_{10-r} = {}^{10}C_r$.Now,consider the ratio $\frac{a_r}{a_{r-

Vedclass · Accepted Answer

$1210$

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(B) Given $a_r$ is the coefficient of $x^{10-r}$ in $(1+x)^{10}$,we have $a_r = {}^{10}C_{10-r} = {}^{10}C_r$.Now,consider the ratio $\frac{a_r}{a_{r-1}} = \frac{{}^{10}C_r}{{}^{10}C_{r-1}} = \frac{10-r+1}{r} = \frac{11-r}{r}$.Substituting this into the sum,we get $\sum \limits_{r=1}^{10} r^3 \left(\frac{11-r}{r}ight)^2 = \sum \limits_{r=1}^{10} r^3 \cdot \frac{(11-r)^2}{r^2} = \sum \limits_{r=1}^{10} r(11-r)^2$.Expanding the term: $\sum \limits_{r=1}^{10} r(121 - 22r + r^2) = \sum \limits_{r=1}^{10} (121r - 22r^2 + r^3)$.Using the summation formulas $\sum r = \frac{n(n+1)}{2}$,$\sum r^2 = \frac{n(n+1)(2n+1)}{6}$,and $\sum r^3 = \left(\frac{n(n+1)}{2}ight)^2$ for $n=10$:$= 121 	imes \frac{10 	imes 11}{2} - 22 	imes \frac{10 	imes 11 	imes 21}{6} + \left(\frac{10 	imes 11}{2}ight)^2$$= 121 	imes 55 - 22 	imes 385 + 3025$$= 6655 - 8470 + 3025 = 1210$.

If $a_r$ is the coefficient of $x^{10-r}$ in the Binomial expansion of $(1+x)^{10}$,then $\sum \limits_{r=1}^{10} r^3\left(\frac{a_r}{a_{r-1}}\right)^2$ is equal to

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