Single Choice

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

A$$12.50$$
B$$12.00$$
C$$12.75$$
D$$12.25$$
Correct Answer

Solution

Given:- $${\left( x + 10 \right)}^{50} + {\left( x - 10 \right)}^{50} = {a}_{0} + {a}_{1} x + {a}_{2} {x}^{2} + ........... + {a}_{50} {x}^{50}$$
To find:- $$\cfrac{{a}_{2}}{{a}_{0}} = \cfrac{\text{coefficient of } {x}^{2}}{\text{Coefficient of } {x}^{0}}$$
As we know that, the general term in an expansion $${\left( a + b \right)}^{n}$$ is given as,
$${T}_{r + 1} = {^{n}{C}_{r}} {\left( a \right)}^{n - r} {\left( b \right)}^{r}$$
Now,
General term of $${\left( x + 10 \right)}^{50}$$-
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 $${\left( x + \left( -10 \right) \right)}^{50}$$-
Here,
$$a = x, b = -10$$
$${T}_{r + 1} = {^{50}{C}_{r}}{\left( x \right)}^{50 - r} {\left( -10 \right)}^{r}$$
For coeficient 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 from the given expansion,
$${a}_{2} = {^{50}{C}_{48}} {\left( 10 \right)}^{48} + {^{50}{C}_{48}} {\left( 10 \right)}^{48} = {^{50}{C}_{48}} \left( {\left( 10 \right)}^{48} + {\left( 10 \right)}^{48} \right)$$
$${a}_{0} = {^{50}{C}_{50}} {\left( 10 \right)}^{50} + {^{50}{C}_{50}} {\left( 10 \right)}^{50} = {^{50}{C}_{50}} \left( {\left( 10 \right)}^{50} + {\left( 10 \right)}^{50} \right)$$
Now,
$$\cfrac{{a}_{2}}{{a}_{0}} = \cfrac{{^{50}{C}_{48}} \left( {\left( 10 \right)}^{48} + {\left( 10 \right)}^{48} \right)}{{^{50}{C}_{50}} \left( {\left( 10 \right)}^{50} + {\left( 10 \right)}^{50} \right)}$$
As we know that,
$${^{n}{C}_{r}} = \cfrac{n!}{r! \left( n - r \right)!}$$
Therefore,
$$\cfrac{{a}_{2}}{{a}_{0}} = \cfrac{50 \times 49}{2} \times \left( \cfrac{{10}^{48}}{{10}^{50}}\right)$$
$$\Rightarrow \cfrac{{a}_{2}}{{a}_{0}} = \cfrac{49}{4}= 12.25$$

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