Question

Let $(x + 10)^{50} + (x- 10)^{50} = a_0 + a_1x + a_2x^2 + .... + a_{50} x^{50}$, for all $x \in R$ ; then $\frac{a_2}{a_0}$ is equal to

Answer 1

$(x + 10)^{50} + (x - 10)^{50}$
$ = a_0 + a_1 x + a_2 x^2 + \equiv + a_{50} x^{50}$
$\therefore a_0 + a_1 x + a_2 x^2 + \equiv + a_{50} x^{50}$
$ = 2(\,{}^{50}C_0 x^{50} + \,{}^{50}C_2 x^{48} . 10^2 + \,{}^{50}C_4 x^{46} . 10^4+ ...)$
$\therefore a_0 = 2 . \,{}^{50} C_{50} 10^{50}$
$a_2 = 2 . \,{}^{50} C_2 . 10^{48}$
$\therefore \frac{a_2}{a_0} = \frac{\,{}^{50}C_2 \times 10^{48}}{\,{}^{50}C_{50} 10^{50}}$
$ = \frac{50 \times 49}{ 2 \times 100} = \frac{49}{4} = 12.25$