Question

Let $(x + 10)^{50} + (x - 10)^{50} = a_0 + a_1 x + a_2x^2 + ..... + a_{50} \; x^{50} , $ for all $x \in R , $ the $\frac{a_2}{a_0} $ is equal to :-

• A. 12.5
• B. 12
• C. 12.75
• D. 12.25

Answer 1

The Correct Option is D

$\left(10+x\right)^{50} +\left(10-x\right)^{50}$
$ \Rightarrow a_{2} = 2. ^{50}C_{2} 10^{48} , a_{0} = 2.10^{50} $
$ \frac{a_{2}}{a_{0}} = \frac{^{50}C_{2}}{10^{2}} = 12.25 $

Answer 2

Given that;
\((x+10)^{50}+(x-10)^{50}=a_0+a_2x^2+......+a_{50}x^{50}\forall{x}\in{R}\)
General term f \((x+10)^{50}\)
Here,
a=x,b=10
\(\therefore T_k+1=^{50}C_k(x)^{50-k}(10)^k\)
For coefficient of \(x^2\)
50-k=2
k=48
\(T_{48+1}=^{50}C_{4_{3}}(x)^{50-48}(10)^{48}\)
For coefficient of \(x^0\)
50-k=0     \(\therefore T_{50+1}=^{50}C_{50}(x)^{50-50}(10)^{50}\)
k=50            ⇒\(T_{52}=^{50}C_{50}(10)^{50}.x^0\)
Now general term of \((x+(-10))^{50}\)
Here, a=x,b=-10
\(T_{k+1}=50_{ck}(x)^{50-k}(-10)^k\)
For coefficient of \(x^2\)
50-k=2
k=48
\(T_{48+1}=^{50}C_{48}(x)^{50-48}(-10)^{48}\)
Similarly for \(x^0\)
\(T_{51}=^{50}C_{50}(10)^{50}x^{0^{1}}\)
\(\therefore a_2=^{50}C_{48}((10)^{48}+(10)^{48})\)
\(a_0=^{50}C_{50}((10)^{50}+(10)^{50})\)
Dividing, \(\frac{a_2}{a_0}=\frac{^{50}C_{48}((10)^{48}+(10)^{48})}{^{50}C_{50}((10)^{50}+(10)^{50})}\)
we know that \(^nC_k=\frac{n!}{k!(n-k)!}\)
\(\therefore \frac{a_2}{a_0}=\frac{50\times49}{2}.\frac{(10^{48})}{10^{50}}\)
⇒\(\frac{a_2}{a_0}=12.25\)