Limit Continuity And Differentiability Ques 1
- If $\left.{ }^{20} C_1+\left(2^2\right){ }^{20} C_2+\left(3^2\right)\right)^{20} C_3+\ldots . .+\left(20^2\right)^{20} C_{20}=A\left(2^\beta\right)$ , then the ordered pair $(A, \beta)$ is equal to
(2019 Main, 12 April II)
(a) $(420,19)$
(b) $(420,18)$
(c) $(380,18)$
(d) $(380,19)$
Show Answer
Answer:
Correct Answer: 1.(b)
Solution: (b) We know,
$ (1+x)^n={ }^n C_0+{ }^n C_1 x+{ }^n C_2 x^2+\ldots+{ }^n C_n x^n $
On differentiating both sides w.r.t. $x$, we get
$ n(1+x)^{n-1}={ }^n C_1+2{ }^n C_2 x+\ldots+n{ }^n C_n x^{n-1} $
On multiplying both sides by $x$, we get
$ n x(1+x)^{n-1}={ }^n C_1 x+2^n C_2 x^2+\ldots+n^n C_n x^n $
Again on differentiating both sides w.r.t. $x$, we get
$ \begin{aligned} n\left[(1+x)^{n-1}\right. \left.+(n-1) x(1+x)^{n-2}\right] \\ ={ }^n C_1+2^2{ }^n C_2 x+\ldots+n^2{ }^n C_n x^{n-1} \end{aligned} $
Now putting $x=1$ in both sides, we get
$ \begin{aligned} &{ }^n C_1+\left(2^2\right){ }^n C_2+\left(3^2\right){ }^n C_3+\ldots+\left(n^2\right){ }^n C_n \\ &=n\left(2^{n-1}+(n-1) 2^{n-2}\right) \end{aligned} $
For $n=20$, we get
$ \begin{aligned} & { }^{20} C_1+\left(2^2\right){ }^{20} C_2+\left(3^2\right){ }^{20} C_3+\ldots+(20)^2{ }^{20} C_{20} \\ & =20\left(2^{19}+(19) 2^{18}\right) \\ & =20(2+19) 2^{18}=420\left(2^{18}\right) \\ & =A\left(2^B\right) \text { (given) } \end{aligned} $
On comparing, we get
$ (A, B)=(420,18) $
BETA
