Question

The term in ${{\left( x+y \right)}^{50}}$ which is greatest in absolute value if $\left| x \right|=\sqrt{3}\left| y \right|$ is
$1)\text{ }{{\text{T}}_{17}}$
$2)\text{ }{{\text{T}}_{19}}$
$3)\text{ }{{\text{T}}_{29}}$
$4)\text{ }{{\text{T}}_{25}}$

Answer 1

Hint: In this question we have been given with a term which is in the form of a simplified polynomial. We will use the binomial expansion formula on the term to get its value given. We have the formula as ${{T}_{r+1}}{{=}^{n}}{{C}_{r}}\times {{x}^{n-r}}\times {{y}^{r}}$. We will use this formula and substitute the value of $n$ as $50$. We will then use the properties of exponents to simplify the exponent. We will then substitute the value of $r$ such that it gives us the highest exponent which is at the middle and we will compare it with the next term to confirm if it is the highest and get our required solution.

Complete step by step solution:
We have the term given to us as:
$\Rightarrow {{\left( x+y \right)}^{50}}$
On using binomial expansion to expand the term, we get:
$\Rightarrow {{T}_{r+1}}=\sum\limits_{r=0}^{50}{^{50}{{C}_{r}}{{\left( x \right)}^{50-r}}{{\left( y \right)}^{r}}}$
Now we know from the question that $x=\sqrt{3}y$ therefore, on substituting, we get:
$\Rightarrow {{T}_{r+1}}=\sum\limits_{r=0}^{50}{^{50}{{C}_{r}}{{\left( \sqrt{3}y \right)}^{50-r}}{{\left( y \right)}^{r}}}$
Now since we have two terms with base as $y$ we can club them together and write it as:
$\Rightarrow {{T}_{r+1}}=\sum\limits_{r=0}^{50}{^{50}{{C}_{r}}{{\left( \sqrt{3} \right)}^{50-r}}{{\left( y \right)}^{r+50-r}}}$
On simplifying the exponent, we get:
$\Rightarrow {{T}_{r+1}}=\sum\limits_{r=0}^{50}{^{50}{{C}_{r}}{{\left( \sqrt{3} \right)}^{50-r}}{{\left( y \right)}^{50}}}$
Now we know that the value of the binomial coefficient is highest at the middle term that is ${{T}_{25}}$ since the expansion starts with $0$, the value of $r$ for the ${{25}^{th}}$ term will be $24$
On substituting $r=24$, we get:
$\Rightarrow {{T}_{24+1}}{{=}^{50}}{{C}_{24}}{{\left( \sqrt{3} \right)}^{50-24}}{{\left( y \right)}^{50}}$
On simplifying the exponent, we get:
$\Rightarrow {{T}_{25}}{{=}^{50}}{{C}_{24}}{{\left( \sqrt{3} \right)}^{26}}{{\left( y \right)}^{50}}$
We will also compare it with the ${{26}^{th}}$ term to compare the value. On taking $r=25$, we get:
$\Rightarrow {{T}_{25+1}}{{=}^{50}}{{C}_{25}}{{\left( \sqrt{3} \right)}^{50-25}}{{\left( y \right)}^{50}}$
On simplifying, we get:
$\Rightarrow {{T}_{26}}{{=}^{50}}{{C}_{26}}{{\left( \sqrt{3} \right)}^{24}}{{\left( y \right)}^{50}}$
On comparing the values of ${{T}_{25}}$ and ${{T}_{26}}$ we get the value of ${{T}_{25}}$ higher therefore, in ${{\left( x+y \right)}^{50}}$the highest value is at ${{T}_{25}}$ given $\left| x \right|=\sqrt{3}\left| y \right|$ therefore, the correct option is $\left( D \right)$.

So, the correct answer is “Option D”.

Note: In this question we have used binomial expansion. Binomial expansion is used to calculate the value of the terms which are in whole square. The general form of expanding the binomial formula should be remembered. Binomial expansion formula should not be confused with binomial distribution formula.