Question

If ${}^{20}{C}_1+(2^2){}^{20}{C}_2+(3^2{)}^{20}{C}_3+........+(20^2{)}^{20}{C}_{20}=A(2^{\beta })$, then the ordered pair $(A,\beta )$ is equal to :

• A. (420, 18)
• B. (380, 19)
• C. (380, 18)
• D. (420, 19)

Answer 1

Answer: (A)

$(1+x{)}^n={}^n{C}_0+{}^n{C}_{1}x+{}^n{C}_2{x}^2+...{+}^n{C}_n{x}^n$
Diff. w.r.t. x
$\Rightarrow n(1+x{)}^{n-1}={}^n{C}_1{+}^n{C}_2(2x)+.....{+}^n{C}_{n}n(x{)}^{n-1}$
Multiply by x both side
$\Rightarrow nx[(1+x{)}^{n-1}{=}^n{C}_{1}x{+}^n{C}_2(2x^2)+.....{+}^n{C}_n(n{x}^n)$
Diff w.r.t. x
$\Rightarrow n[(1+x{)}^{n-1}+(n-1])x(1+x{)}^{n-2}$
${}^n{C}_1{+}^n{C}_2{2}^{2}x+....{.}^n{C}_n(n^2)x^{n-1}$
Put $x=1$ and $n=20$
$\Rightarrow {}^{20}{C}_1+2^2{}^{20}{C}_2+3^2{}^{20}{C}_3+.....(20{)}^2{}^{20}{C}_{20}$
= 20 $\times 2^{18}[2+19]=420(2^{18})A(2^{\beta })$