Let (x+10)^50+(x-10)^50=a_0+a_1x+a_2x^2++a_50x^50 for all x R; then a_2a_0 is equal to:

Question

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

Accepted Answer

$$(x+10)^{50}+(x-10)^{50}=2\left[{}^{50}C_0x^{50}+{}^{50}C_2x^{48}\cdot10^2+{}^{50}C_4x^{46}\cdot10^4+\cdots\right]$$

$a_0=2\cdot{}^{50}C_{50}\cdot10^{50}$, $a_2=2\cdot{}^{50}C_2\cdot10^{48}$

$$\frac{a_2}{a_0}=\frac{{}^{50}C_2}{{}^{50}C_{50}\times10^2}=\frac{50\times49/2}{100}=\frac{49}{4}=12.25$$

Solutions · 1