# Solve the following systems of equations:

Question:

If ${ }^{20} \mathrm{C}_{1}+\left(2^{2}\right)^{20} \mathrm{C}_{2}+\left(3^{2}\right)^{20} \mathrm{C}_{3}+\ldots \ldots+\left(20^{2}\right)^{20} \mathrm{C}_{20}$ $=\mathrm{A}\left(2^{\beta}\right)$, then the ordered pair $(A, \beta)$ is equal to:

1. $(420,18)$

2. $(380,19)$

3. $(380,18)$

4. $(420,19)$

Correct Option: 1

Solution:

$(1+x)^{n}={ }^{n} C_{0}+{ }^{n} C_{1} x+{ }^{n} C_{2} x^{2}+\ldots .+{ }^{n} C_{n} x^{n}$

Diff. w.r.t. $x$

$\Rightarrow \mathrm{n}(1+\mathrm{x})^{\mathrm{n}-1}=\mathrm{n}_{1}+\mathrm{n}_{2}(2 \mathrm{x})+\ldots \ldots+\mathrm{n}_{\mathrm{n}} \mathrm{n}(\mathrm{x})^{\mathrm{n}-1}$

Multiply by x both side

$\Rightarrow n x(1+x)^{n-1}=n_{1} x+n C_{2}\left(2 x^{2}\right)+\ldots .+n_{n}\left(n x^{n}\right)$

Diff w.r.t. $\mathrm{x}$

$\Rightarrow \mathrm{n}\left[(1+\mathrm{x})^{\mathrm{n}-1}+(\mathrm{n}-1) \mathrm{x}(1+\mathrm{x})^{\mathrm{n}-2}\right]$

$={ }^{n} C_{1}+{ }^{n} C_{2} 2{ }^{2} x+\ldots .{ }^{n} C_{n}\left(n^{2}\right) x^{n-1}$

Put $\mathrm{x}=1$ and $\mathrm{n}=20$

$\Rightarrow{ }^{20} \mathrm{C}_{1}+2^{2}{ }^{20} \mathrm{C}_{2}+3^{2}{ }^{20} \mathrm{C}_{3}+\ldots .+(20)^{2}{ }^{20} \mathrm{C}_{20}$

$=20 \times 2^{18}[2+19]=420\left(2^{18}\right)=\mathrm{A}\left(2^{\beta}\right)$