Question

Let $(x+10{)}^{50}+(x-10{)}^{50}=a_0+a_{1}x+a_2{x}^2+....+a_{50}{x}^{50}$, for all $x\in R$ ; then $\frac{a_2}{a_0}$ is equal to

Answer 1

Answer: 12.25

$(x+10{)}^{50}+(x-10{)}^{50}$
$=a_0+a_{1}x+a_2{x}^2+\equiv +a_{50}{x}^{50}$
$\therefore a_0+a_{1}x+a_2{x}^2+\equiv +a_{50}{x}^{50}$
$=2({}^{50}{C}_0{x}^{50}+{}^{50}{C}_2{x}^{48}.10^2+{}^{50}{C}_4{x}^{46}.10^4+...)$
$\therefore a_0=2.{}^{50}{C}_{50}10^{50}$
$a_2=2.{}^{50}{C}_2.10^{48}$
$\therefore \frac{a_2}{a_0}=\frac{{}^{50}{C}_2\times 10^{48}}{{}^{50}{C}_{50}10^{50}}$
$=\frac{50\times 49}{2\times 100}=\frac{49}{4}=12.25$