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:

• A. $12.75$
• B. $12.50$
• C. $12.25$
• D. $12.00$

Answer 1

The correct option is C $12.25$

$(x+10{)}^{50}={}^{50}{C}_0(10{)}^{50}+{}^{50}{C}_{1}x\cdot {10}^{49}+{}^{50}{C}_2{x}^2\cdot {10}^{48}+\cdots \left(1\right)(x-10{)}^{50}={}^{50}{C}_0(10{)}^{50}-{}^{50}{C}_{1}x\cdot {10}^{49}+{}^{50}{C}_2{x}^2\cdot {10}^{48}-\cdots \left(2\right)$

Add equation $\left(1\right)$ and $\left(2\right)$
$(x+10{)}^{50}+(x-10{)}^{50}=2\cdot {}^{50}{C}_0\cdot (10{)}^{50}+2\cdot {}^{50}{C}_2(10{)}^{48}\cdot x^2+\cdots \therefore a_0=2\cdot {}^{50}{C}_0(10{)}^{50},a_2=2\cdot {}^{50}{C}_2(10{)}^{48}\Rightarrow \frac{a_2}{a_0}=\frac{2\cdot {}^{50}{C}_2(10{)}^{48}}{2\cdot {}^{50}{C}_0(10{)}^{50}}\therefore \frac{a_2}{a_0}=\frac{50\times 49}{2\times 100}=12.25$