Question

If ${}^{20}{C_1} + \left( {{2^2}} \right){}^{20}{C_2} + \left( {{3^2}} \right){}^{20}{C_3} + \ldots + \left( {{{20}^2}} \right){}^{20}{C_{20}} = A\left( {{2^\beta }} \right)$, then the ordered pair $\left( {A,\beta } \right)$ is equal to
A. $\left( {420,18} \right)$
B. $\left( {380,19} \right)$
C. $\left( {380,18} \right)$
D. $\left( {420,19} \right)$

Answer 1

In this problem, first we need to expand ${\left( {1 + x} \right)^n}$ using the binomial expansion. Now differentiate the binomial expansion with respect to$x$. Next, multiply with $x$ on both sides. Further, differentiate the obtained expression with respect to$x$. Substitute, $x = 1$ and $n=20$ in obtained expression.

Complete step-by-step solution:
The binomial expansion of the function ${\left( {1 + x} \right)^n}$ is shown below.
${\left( {1 + x} \right)^n} = {}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + \ldots + {}^n{C_n}{x^n}$
Differentiate, the above binomial expansion with respect to$x$.
n{\left( {1 + x} \right)^{n - 1}} = 0 + {}^n{C_1}\left( 1 \right) + {}^n{C_2}\left( {2x} \right) + \ldots + {}^n{C_n}\left( {n{x^{n - 1}}} \right) \\\
$\Rightarrow n{\left( {1 + x} \right)^{n - 1}} = {}^n{C_1} + {}^n{C_2}\left( {2x} \right) + \ldots + {}^n{C_n}\left( {n{x^{n - 1}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 1 \right) \\\$
Multiply with $x$ on both sides of the equation (1).
nx{\left( {1 + x} \right)^{n - 1}} = {}^n{C_1}x + {}^n{C_2}\left( {2x} \right)\left( x \right) + \ldots + {}^n{C_n}\left( {n{x^{n - 1}}} \right)\left( x \right) \\\
$\Rightarrow nx{\left( {1 + x} \right)^{n - 1}} = {}^n{C_1}x + {}^n{C_2}\left( {2{x^2}} \right) + \ldots + {}^n{C_n}\left( {n{x^n}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 2 \right) \\\$
Further, differentiate equation (2) with respect to $x$ as shown below.
n\left[ {{{\left( {1 + x} \right)}^{n - 1}} + \left( {n - 1} \right)x{{\left( {1 + x} \right)}^{n - 2}}} \right] = {}^n{C_1}\left( 1 \right) + {}^n{C_2}\left( {2 \cdot 2x} \right) + \ldots + {}^n{C_n}\left( {n \cdot n{x^{n - 1}}} \right) \\\
$\Rightarrow n\left[ {{{\left( {1 + x} \right)}^{n - 1}} + \left( {n - 1} \right)x{{\left( {1 + x} \right)}^{n - 2}}} \right] = {}^n{C_1}\left( 1 \right) + {}^n{C_2}\left( {{2^2}x} \right) + \ldots + {}^n{C_n}\left( {{n^2}{x^{n - 1}}} \right)$
Substitute, 1 for $x$ and 20 for $n$ in above expression.
20\left[ {{{\left( {1 + 1} \right)}^{20 - 1}} + \left( {20 - 1} \right)\left( 1 \right){{\left( {1 + 1} \right)}^{20 - 2}}} \right] = {}^{20}{C_1} + {}^{20}{C_2}\left( {{2^2}\left( 1 \right)} \right) + \ldots + {}^{20}{C_{20}}\left( {{{20}^2}{{\left( 1 \right)}^{20 - 1}}} \right) \\\
\Rightarrow 20\left[ {{{\left( 2 \right)}^{19}} + 19{{\left( 2 \right)}^{18}}} \right] = {}^{20}{C_1} + \left( {{2^2}} \right){}^{20}{C_2} + \ldots + \left( {{{20}^2}} \right){}^{20}{C_{20}} \\\
\Rightarrow 20 \cdot {2^{18}}\left[ {2 + 19} \right] = {}^{20}{C_1} + \left( {{2^2}} \right){}^{20}{C_2} + \ldots + \left( {{{20}^2}} \right){}^{20}{C_{20}} \\\
$\Rightarrow 420\left( {{2^{18}}} \right) = {}^{20}{C_1} + \left( {{2^2}} \right){}^{20}{C_2} + \ldots + \left( {{{20}^2}} \right){}^{20}{C_{20}}$
Now, compare $420\left( {{2^{18}}} \right)$ with $A\left( {{2^\beta }} \right)$ to obtain the value of $A$ and$\beta$.
A = 420 \\\
$\Rightarrow \beta = 18$
Thus, the ordered pair $\left( {A,\beta } \right)$ is equal to$\left( {420,18} \right)$, hence, option (A) is correct answer.

Note: Binomial expansion is expansion of the powers of binomials or sum of two terms with the help of the combinations. In this problem, after obtaining the second derivative of the function ${\left( {1 + x} \right)^n}$ we need to substitute 1 for x to obtain the value of given ordered pair.