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,
$$ \begin{gathered} \quad(1+x)^{20}={ }^{20} C _0+{ }^{20} C _1 x+{ }^{20} C _2 x^{2}+\ldots+ \\ { }^{20} C _{r-1} x^{r-1}+{ }^{20} C _r x^{r}+\ldots+{ }^{20} C _{20} x^{20} \\ \therefore \quad(1+x)^{20} \cdot(1+x)^{20}=\left({ }^{20} C _0+{ }^{20} C _1 x+\right. \\ \left.{ }^{20} C _2 x^{2}+\ldots+{ }^{20} C _{r-1} x^{r-1}+{ }^{20} C _r x^{r}+\ldots+{ }^{20} c _{20} x^{20}\right) \\ \times\left({ }^{20} C _0+{ }^{20} C _1 x+\ldots+{ }^{20} C _{r-1} x^{r-1}+{ }^{20} C _r x^{r}\right. \\ \left.+\ldots+{ }^{20} C _{20} x^{20}\right) \\ \Rightarrow(1+x){ }^{40}=\left({ }^{20} C _0 \cdot{ }^{20} C _r+{ }^{20} C _1{ }^{20} C _{r-1} \ldots\right. \\ \left.{ }^{20} C _r{ }^{20} C _0\right) x^{r}+\ldots \end{gathered} $$
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 .
