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 the value of the $\dfrac{{{a_2}}}{{{a_0}}}$
 (A) 12.50
 (B) 12.00
 (C) 12.75
 (D) 12.25

Answer 1

Hint: In this question, the term ${a_2}$ = coefficient of ${x^2}$ and the term ${a_0}$ = coefficient of ${x^0}$, then you can find the value of $\dfrac{{{a_2}}}{{{a_0}}} $= $\dfrac{{{\text{coefficient of }}{x^2}}}{{{\text{coefficient of }}{x^0}}}$.

Complete step-by-step solution -
As we know that, the general term in the expansion ${(a + b)^n}$ is given by
\[{t_{r + 1}} = {}^n{C_r}{(a)^{n - r}}{(b)^r}.............(1)\]
Now, the given expansion ${\left( {x + 10} \right)^{50}}$ is comparing with the expansion ${\left( {a + b} \right)^n}$ and then we have
$a = x, b = 10{\text{ and }}n = 50$
Now put all the values a, b and n in the equation (1), we get
\[{t_{r + 1}} = {}^{50}{C_r}{(x)^{50 - r}}{(10)^r}.............(2)\]

For coefficient of ${x^2}$, we must have
$50 - r = 2$
$r = 50 - 2$
$r = 48$
Put the value of $r = 48$ in the equation (2), we get
\[{t_{48 + 1}} = {}^{50}{C_{48}}{(x)^{50 - 48}}{(10)^{48}}\]
\[{t_{49}} = {}^{50}{C_{48}}{(x)^2}{(10)^{48}}\]
\[{t_{49}} = {}^{50}{C_{48}}{(10)^{48}}{x^2}...............(3)\]

For coefficient of ${x^0}$, we must have
$50 - r = 0$
$r = 50$
Put the value of $r = 50$ in the equation (2), we get
\[{t_{50 + 1}} = {}^{50}{C_{50}}{(x)^{50 - 50}}{(10)^{50}}\]
\[{t_{51}} = {}^{50}{C_{50}}{(x)^0}{(10)^{50}}\]
\[{t_{51}} = {}^{50}{C_{50}}{(10)^{50}}{x^0}................(4)\]
Also, the given second expansion ${\left( {x - 10} \right)^{50}} = {\left[ {x + ( - 10)} \right]^{50}}$ is comparing with the expansion ${\left( {a + b} \right)^n}$ and then we have
$a = x, b = - 10{\text{ and }}n = 50$
Now put all the values a, b and n in the equation (1), we get
\[{t_{r + 1}} = {}^{50}{C_r}{(x)^{50 - r}}{( - 10)^r}.............(5)\]

For coefficient of ${x^2}$, we must have
$50 - r = 2$
$r = 50 - 2$
$r = 48$
Put the value of $r = 48$ in the equation (5), we get
\[{t_{48 + 1}} = {}^{50}{C_{48}}{(x)^{50 - 48}}{( - 10)^{48}}\]
\[{t_{49}} = {}^{50}{C_{48}}{(x)^2}{(10)^{48}}\]
\[{t_{49}} = {}^{50}{C_{48}}{(10)^{48}}{x^2}.............(6)\]

For coefficient of ${x^0}$, we must have
$50 - r = 0$
$r = 50$
Put the value of $r = 50$ in the equation (5), we get
\[{t_{50 + 1}} = {}^{50}{C_{50}}{(x)^{50 - 50}}{( - 10)^{50}}\]
\[{t_{51}} = {}^{50}{C_{50}}{(x)^0}{(10)^{50}}\]
\[{t_{51}} = {}^{50}{C_{50}}{(10)^{50}}{x^0}.............(7)\]
From the equations (3) and (6), we get
Coefficient of ${x^2} = {a_2} = {}^{50}{C_{48}}{(10)^{48}} + {}^{50}{C_{48}}{(10)^{48}} = {}^{50}{C_{48}}\left[ {{{(10)}^{48}} + {{(10)}^{48}}} \right]$
From the equations (4) and (7), we get
Coefficient of ${x^0} = {a_0} = {}^{50}{C_{50}}{(10)^{50}} + {}^{50}{C_{50}}{(10)^{50}} = {}^{50}{C_{50}}\left[ {{{(10)}^{50}} + {{(10)}^{50}}} \right]$
Consider the ratio of coefficient of ${x^2}$ and coefficient of ${x^0}$,

\[\dfrac{{{a_2}}}{{{a_0}}} = \dfrac{{{}^{50}{C_{48}}\left[ {{{(10)}^{48}} + {{(10)}^{48}}} \right]}}{{{}^{50}{C_{50}}\left[ {{{(10)}^{50}} + {{(10)}^{50}}} \right]}} = \dfrac{{{}^{50}{C_{48}}}}{{{}^{50}{C_{50}}}} \times \dfrac{{2 \times {{(10)}^{48}}}}{{2 \times {{(10)}^{50}}}} = \dfrac{{{}^{50}{C_{48}}}}{{{}^{50}{C_{50}}}} \times \dfrac{{{{(10)}^{48}}}}{{{{(10)}^{50}}}} = \dfrac{{{}^{50}{C_{48}}}}{{{}^{50}{C_{50}}}} \times \dfrac{1}{{{{(10)}^2}}}.......(8)\]

By using the formula of combination, ${}^n{C_r} = \dfrac {{n!}}{{r!(n - r)!}}$ .

We have ${}^{50}{C_{48}}$ =$ \dfrac{{50!}}{{48!(2!)}}$ = $\dfrac{{50 \times 49}}{{1 \times 2}} $ and $ {}^{50}{C_{50}} = \dfrac{{50!}}{{50!(0!)}} = 1 $
Put the values of $^{50}{C_{48}}$ and $^{50}{C_{50}}$ in the equation (8), we get
\[\dfrac{{{a_2}}}{{{a_0}}} = \dfrac{{50 \times 49}}{2} \times \dfrac{1}{{{{(10)}^2}}}\]
\[\dfrac{{{a_2}}}{{{a_0}}} = \dfrac{{50 \times 49}}{2} \times \dfrac{1}{{100}} = \dfrac{{49}}{{2 \times 2}} = \dfrac{{49}}{4} = 12.25\]
Hence, the correct option of the given question is option (d).

Note: Alternatively this question is solved by using the formula \[{(x + a)^n} + {(x - a)^n} = 2\left[ {^n{C_0}{x^n}{a^0}{ + ^{n - 2}}{C_0}{x^{n - 2}}{a^2}{ + ^{n - 4}}{C_0}{x^{n - 4}}{a^4} + .............} \right]\]. As the terms having odd power of a will be canceled out and the terms of having even power of a will be added.