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

Answer 1

Answer: (D)

$12.25$

Given:- ${(x+10)}^{50}+{(x-10)}^{50}=a_0+a_{1}x+a_2{x}^2+...........+a_{50}{x}^{50}$

To find:- $\frac{a_2}{a_0}=\frac{\text{coefficient of}{x}^2}{\text{Coefficient of}{x}^0}$

As we know that, the general term in an expansion ${(a+b)}^n$ is given as,

$T_{r+1}={}^n{C}_r{\left(a\right)}^{n-r}{\left(b\right)}^r$

Now,

General term of ${(x+10)}^{50}$-

Here,

$a=x,b=10$

$T_{r+1}={}^{50}{C}_r{\left(x\right)}^{50-r}{\left(10\right)}^r$

For coefficient of $x^2$-

$50-r=2\Rightarrow r=48$

$T_{48+1}={}^{50}{C}_{48}{\left(x\right)}^{50-48}{\left(10\right)}^{48}$

$T_{49}={}^{50}{C}_{48}{\left(10\right)}^{48}{x}^2$

For coeficient of $x^0$-

$50-r=0\Rightarrow r=50$

$T_{50+1}={}^{50}{C}_{50}{\left(x\right)}^{50-50}{\left(10\right)}^{50}$

$T_{51}={}^{50}{C}_{50}{\left(10\right)}^{50}{x}^0$

Now,

General term of ${(x+(-10))}^{50}$-

Here,

$a=x,b=-10$

$T_{r+1}={}^{50}{C}_r{\left(x\right)}^{50-r}{(-10)}^r$

For coeficient of $x^2$-

$50-r=2\Rightarrow r=48$

$T_{48+1}={}^{50}{C}_{48}{\left(x\right)}^{50-48}{(-10)}^{48}$

$T_{49}={}^{50}{C}_{48}{\left(10\right)}^{48}{x}^2$

For coeficient of $x^0$-

$50-r=0\Rightarrow r=50$

$T_{50+1}={}^{50}{C}_{50}{\left(x\right)}^{50-50}{(-10)}^{50}$

$T_{51}={}^{50}{C}_{50}{\left(10\right)}^{50}{x}^0$

Now from the given expansion,

$a_2={}^{50}{C}_{48}{\left(10\right)}^{48}+{}^{50}{C}_{48}{\left(10\right)}^{48}={}^{50}{C}_{48}({\left(10\right)}^{48}+{\left(10\right)}^{48})$

$a_0={}^{50}{C}_{50}{\left(10\right)}^{50}+{}^{50}{C}_{50}{\left(10\right)}^{50}={}^{50}{C}_{50}({\left(10\right)}^{50}+{\left(10\right)}^{50})$

Now,

$\frac{a_2}{a_0}=\frac{{}^{50}{C}_{48}({\left(10\right)}^{48}+{\left(10\right)}^{48})}{{}^{50}{C}_{50}({\left(10\right)}^{50}+{\left(10\right)}^{50})}$

As we know that,

${}^n{C}_r=\frac{n!}{r!(n-r)!}$

Therefore,

$\frac{a_2}{a_0}=\frac{50\times 49}{2}\times \left(\frac{{10}^{48}}{{10}^{50}}\right)$

$\Rightarrow \frac{a_2}{a_0}=\frac{49}{4}=12.25$