Let $a_r=r^2\frac{{ }^{50}{C}_r}{{ }^{50}{C}_{r-1}}$ and $b$ denote the coefficient of $x^{49}$ in $\left(x-a_1\right)\left(x-a_2\right)\dots \left(x-a_{50}\right),$ then $\frac{\left|b\right|}{650}$ is k then k-29=

We have,

$a_r=r^2\frac{{ }^{50}{C}_r}{{ }^{50}{C}_{r-1}}=r^2\left(\frac{50-r+1}{r}\right)$

It is known that,

$\frac{{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}}{{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}-1}}=\frac{\mathrm{n}-\mathrm{r}+1}{\mathrm{r}}{\mathrm{a}}_{\mathrm{r}}=\mathrm{r}(51-\mathrm{r})=51\mathrm{r}-{\mathrm{r}}^2$

here, b =coefficient of $x^{49}\text{ in }\left(x-a_1\right)\left(x-a_2\right)\left(x-a_3\right)\dots \dots ..\left(x-a_{49}\right)\left(x-a_{50}\right)$

So,

$\begin{array}{l}\mathrm{b}=-\left({\mathrm{a}}_1+{\mathrm{a}}_2+{\mathrm{a}}_3+\dots \dots \dots \dots {\mathrm{a}}_{49}+{\mathrm{a}}_{50}\right)\\ \mathrm{b}=-\sum _{\mathrm{r}=1}^{50} {\mathrm{a}}_{\mathrm{r}}=-\sum _{\mathrm{r}=1}^{50} 51\mathrm{r}-{\mathrm{r}}^2=-\left[\left(\frac{51\times 50\times 51}{2}\right)-\left(\frac{50\times 51\times 101}{6}\right)\right]\end{array}$Now, we have,

$$\begin{gathered} \therefore b=-(65025-42925) \\ \Rightarrow b=-22100 \end{gathered}$$

Thus,

$$\begin{gathered} \frac{\left|b\right|}{650}=\frac{|-22100|}{650} \\ \frac{\left|b\right|}{650}=34 \end{gathered}$$