Binomial Theorem 2 Question 2
2.
The value of $r$ for which
${ }^{20} C_{r}{ }^{20} C_{0}+{ }^{20} C_{r-1}{ }^{20} C_{1}+{ }^{20} C_{r-2}{ }^{20} C_{2}+\ldots .+{ }^{20} C_{0}{ }^{20} C_{r}$ is maximum, is
(2019 Main, 11 Jan I)
(a) $15$
(b) $10$
(c) $11$
(d) $20$
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Answer:
Correct Answer: 2. (d)
Solution:
- We know that,
$\quad(1+x)^{20}={ }^{20} C_{0}+{ }^{20} C_{1} x+{ }^{20} C_{2} x^{2}+\ldots+ $ ${ }^{20} C_{r-1} x^{r-1}+{ }^{20} C_{r} x^{r}+\ldots+{ }^{20} C_{20} x^{20} $
$\therefore \quad(1+x)^{20} \cdot(1+x)^{20}=\left({ }^{20} C_{0}+{ }^{20} C_{1} x+\right. $ $\left.{ }^{20} C_{2} x^{2}+\ldots+{ }^{20} C_{r-1} x^{r-1}+{ }^{20} C_{r} x^{r}+\ldots+{ }^{20} c_{20} x^{20}\right) $
$\times\left({ }^{20} C_{0}+{ }^{20} C_{1} x+\ldots+{ }^{20} C_{r-1} x^{r-1}+{ }^{20} C_{r} x^{r}\right. $ $\left.+\ldots+{ }^{20} C_{20} x^{20}\right) $
$\Rightarrow(1+x){ }^{40}=\left({ }^{20} C_{0} \cdot{ }^{20} C_{r}+{ }^{20} C_{1}{ }^{20} C_{r-1} \ldots\right. $ $\left.{ }^{20} C_{r}{ }^{20} C_{0}\right) x^{r}+\ldots$
On comparing the coefficient of $x^{r}$ of both sides, we get
$ { }^{20} C_{0}{ }^{20} C_{r}+{ }^{20} C_{1}{ }^{20} C_{r-1}+\ldots+{ }^{20} C_{r}{ }^{20} C_{0}={ }^{40} C_{r} $
The maximum value of ${ }^{40} C_{r}$ is possible only when $r=20$
$ \left[\because{ }^{n} C_{n / 2} \text { is maximum when } n \text { is even }\right] $
Thus, required value of $r$ is $20$ .
