Question

the coefficients of $x^{49}$ in the polynomial.
$(x-\frac{C_1}{C_0})(x-2^2\frac{C_2}{C_1})(x-3^2\frac{C_3}{C_2}).....(x-{50}^2\frac{C_{50}}{C_{49}})$ is

• A. $50\sum _{r=1}^{50}r-r^2$
• B. $51\sum _{r=1}^{50}{r}^2-\sum _{r=1}^{50}r$
• C. $51\sum _{r=1}^{50}r-\sum _{r=1}^{50}{r}^2$
• D. $noneofthese$

Answer 1

The correct option is A $50\sum _{r=1}^{50}r-r^2$

We Know that $(x-a)(x-b)(x-c).....50$
factors $=x^{50}{x}^{49}.S_1+x^{48}{S}_2-....$
The coefficient of $x^{49}$ is
$S_1=-[\frac{C_1}{C_0}+2^2\frac{C_2}{C_1}+3^2\frac{C_3}{C_2}]+.....r^2\frac{C_r}{C_{r-1}}]$
$Now{r}^2.\frac{C_r}{C_{r-1}}=r^2.\frac{n-r+1}{r}$
$=r(51-r)=51r-r^2$
$\therefore {\sum }_{r=1}^{50}{r}^2\frac{c_r}{C_{r-1}}51{\sum }_{r=1}^{50}r-{\sum }_{r=1}^{50}{r}^2$
$=51.\frac{N(N+1)}{2}-\frac{N(N+1)(2N+1)}{6}$
$=51.\frac{50.51)}{2}-\frac{50.51.\left(101\right)}{6}$
$=\frac{1}{6}50.51[3\times 51-101]=25\times 17\times 52$