Question

If $f(x)={}^{40}{C}_1\cdot x(1-x{)}^{39}+2\cdot {}^{40}{C}_2{x}^2(1-x{)}^{38}+3^{40}{C}_3\times$ $x^3(1-x{)}^{37}+\dots +40\cdot {}^{40}{C}_{40}{x}^{40},$ then the value of $f(3)$ is not divisible by

• A. 5
• B. 6
• C. 8
• D. None of these

Answer 1

Answer: (D)

$f(x)={}^{40}{C}_1\cdot x(1-x{)}^{39}+2\cdot {}^{40}{C}_2{x}^2(1-x{)}^{38}$
$+3{}^{40}{C}_3{x}^3(1-x{)}^{37}+\dots +40\cdot {}^{40}{C}_{40}{x}^{40}$
$T_r=r\cdot {}^{40}{C}_r{x}^r(1-x{)}^{40-r}=40x\cdot {}^{39}{C}_{r-1}{x}^{r-1}(1-x{)}^{40-r}$
$\therefore f(x)=40x(x+1-x{)}^{39}=40x$