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  4. If 20C1 + (22) 20C2 + (32) 20C3 + ........ + (202) 20C20 = A(2β), then the ordered pair (A, β) is equal to :

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Q. 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 :

JEE MainJEE Main 2019Binomial Theorem

Solution:

$(1+x)^n \, = \,{}^{n}C_{0} \, + \,{}^{n}C_{1}x \, + \,{}^{n}C_{2} x^2 + ... + ^{n}C_{n}x^n$
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} 2^2x + ..... ^{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 $\times \, 2^{18} [2 + 19] \, = \, 420 (2^{18}) \, A(2^{\beta})$