Question

The value of $r$ for which
${}^{20}{{C}_{r}}{}^{20}{{C}_{0}}+{}^{20}{{C}_{r-1}}{}^{20}{{C}_{1}}+{}^{20}{{C}_{r-2}}{}^{20}{{C}_{2}}+---+{}^{20}{{C}_{0}}{}^{20}{{C}_{r}}$
is maximum is
(a) $20$
(b) $15$
(c) $11$
(d) $10$

Answer 1

We have to find the value of $r$ at which ${}^{20}{{C}_{r}}{}^{20}{{C}_{0}}+---+{}^{20}{{C}_{0}}{}^{20}{{C}_{r}}$ is maximum. To do so, we will first use the Binomial expansion of $n$ which is given as, ${{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{o}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+--+{}^{n}{{C}_{n}}{{x}^{n}}$ with this we will expand ${{\left( 1+x \right)}^{20}}$ and ${{\left( 1+x \right)}^{40}}$ as we know ${{x}^{a}}.{{x}^{b}}={{x}^{a+b}}$ so ${{\left( 1+x \right)}^{20}}.{{\left( 1+x \right)}^{20}}={{\left( 1+x \right)}^{40}}$ , we will use this to expand ${{\left( 1+x \right)}^{40}}$ . Using this, we will get the general term of ${{\left( 1+x \right)}^{40}}$ . And we know for $n=even$ ${{T}_{r}}$ is maximum at $r=\dfrac{n}{2}$ , we will use this to get the final answer.

Complete step-by-step answer:
We are asked to find the value of $r$ such that
${}^{20}{{C}_{r}}{}^{20}{{C}_{0}}+{}^{20}{{C}_{r-1}}{}^{20}{{C}_{1}}+{}^{20}{{C}_{r-2}}{}^{20}{{C}_{2}}+---+{}^{20}{{C}_{0}}{}^{20}{{C}_{r}}$ is maximum.
We know that Binomial expansion of ${{\left( 1+n \right)}^{n}}$ is given as,
${{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+--+{}^{n}{{C}_{n}}{{x}^{n}}$
So for $n=20$
${{\left( 1+x \right)}^{20}}$ will be given as,
${{\left( 1+x \right)}^{20}}={}^{20}{{C}_{0}}+{}^{20}{{C}_{1}}x+{}^{20}{{C}_{2}}{{x}^{2}}+--+{}^{20}{{C}_{20}}{{x}^{20}}$
Similarly ${{\left( 1+x \right)}^{40}}$ will be written as,
${{\left( 1+x \right)}^{40}}={}^{40}{{C}_{0}}+{}^{40}{{C}_{1}}x+--+{}^{40}{{C}_{40}}{{x}^{40}}$
Now we know,
${{\left( 1+x \right)}^{40}}={{\left( 1+x \right)}^{20+20}}$ $\left[ as\ {{x}^{a+b}}={{x}^{a}}.{{x}^{b}} \right]$
So
$\text{ }={{\left( 1+x \right)}^{20}}{{\left( 1+x \right)}^{20}}$
Now putting value of ${{\left( 1+x \right)}^{20}}$ from above expression, we get,
${{\left( 1+x \right)}^{40}}=\left( {}^{20}{{C}_{0}}+{}^{20}{{C}_{1}}x+--+{}^{20}{{C}_{20}}{{x}^{20}} \right)\left( {}^{20}{{C}_{0}}+{}^{20}{{C}_{1}}x+--+{}^{20}{{C}_{20}}{{x}^{20}} \right)$
When we expand the general term for ${{\left( 1+x \right)}^{40}}$ will be given as, ${{T}_{r}}={}^{20}{{C}_{0}}{}^{20}{{C}_{r}}{{x}^{r}}+{}^{20}{{C}_{1}}{{x}^{r}}.{}^{20}{{C}_{r-1}}+---+{}^{20}{{C}_{r}}{{x}^{r}}.{}^{20}{{C}_{0}}$
Taking ${{x}^{r}}$ common , we get,
${{T}_{r}}=\left( {}^{20}{{C}_{0}}{}^{20}{{C}_{r}}+{}^{20}{{C}_{1}}{}^{20}{{C}_{r-1}}+---+{}^{20}{{C}_{r}}{}^{20}{{C}_{0}} \right){{x}^{r}}$
Now comparing the general with general term of ${{\left( 1+x \right)}^{40}}$ which is given as $^{40}{{C}_{r}}{{x}^{r}}$ we get,
$\left( {}^{20}{{C}_{0}}{}^{20}{{C}_{r}}+{}^{20}{{C}_{1}}{}^{20}{{C}_{r-1}}+---+{}^{20}{{C}_{r}}{}^{20}{{C}_{0}} \right){{x}^{r}}{{=}^{40}}{{C}_{r}}{{x}^{r}}$
${{\Rightarrow }^{40}}{{C}_{r}}={}^{20}{{C}_{0}}{}^{20}{{C}_{r}}+{}^{20}{{C}_{1}}{}^{20}{{C}_{r-1}}+---+{}^{20}{{C}_{r}}{}^{20}{{C}_{0}}$
So maximum value of ${}^{20}{{C}_{0}}{}^{20}{{C}_{r}}+{}^{20}{{C}_{1}}{}^{20}{{C}_{r-1}}+---+{}^{20}{{C}_{r}}{}^{20}{{C}_{0}}$ in same as maximum value of $^{40}{{C}_{r}}$ .
We know that,
${}^{n}{{C}_{r}}$ is maximum when $r=\dfrac{n}{2}$
If $n$ is even,
For $^{40}{{C}_{r}}$ , $n=40$ which is even. So we get,
$^{40}{{C}_{r}}$ is maximum at $r=\dfrac{n}{2}$
$\Rightarrow r=\dfrac{40}{2}=20$

So, the correct answer is “Option (a)”.

Note: When we multiply two terms with more than $1$ element then each element of first term get multiplied by other terms all elements , this same process will be done when ${{\left( 1+x \right)}^{20}}$ is multiplied with ${{\left( 1+x \right)}^{20}}$ with the help of this we get the general term ${{T}_{r}}$ for ${{\left( 1+x \right)}^{40}}$ .