Solveeit Logo
Login

Question

Question: The term in \({{\left( x+y \right)}^{50}}\) which is greatest in absolute value if \(\left| x \right...

The term in (x+y)50{{\left( x+y \right)}^{50}} which is greatest in absolute value if ∣x∣=3∣y∣\left| x \right|=\sqrt{3}\left| y \right| is
1) T171)\text{ }{{\text{T}}_{17}}
2) T192)\text{ }{{\text{T}}_{19}}
3) T293)\text{ }{{\text{T}}_{29}}
4) T254)\text{ }{{\text{T}}_{25}}

Explanation

Solution

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 Tr+1=nCr×xn−r×yr{{T}_{r+1}}{{=}^{n}}{{C}_{r}}\times {{x}^{n-r}}\times {{y}^{r}}. We will use this formula and substitute the value of nn as 5050. We will then use the properties of exponents to simplify the exponent. We will then substitute the value of rr 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:
⇒(x+y)50\Rightarrow {{\left( x+y \right)}^{50}}
On using binomial expansion to expand the term, we get:
⇒Tr+1=∑r=05050Cr(x)50−r(y)r\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=3yx=\sqrt{3}y therefore, on substituting, we get:
⇒Tr+1=∑r=05050Cr(3y)50−r(y)r\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 yy we can club them together and write it as:
⇒Tr+1=∑r=05050Cr(3)50−r(y)r+50−r\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:
⇒Tr+1=∑r=05050Cr(3)50−r(y)50\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 T25{{T}_{25}} since the expansion starts with 00, the value of rr for the 25th{{25}^{th}} term will be 2424
On substituting r=24r=24, we get:
⇒T24+1=50C24(3)50−24(y)50\Rightarrow {{T}_{24+1}}{{=}^{50}}{{C}_{24}}{{\left( \sqrt{3} \right)}^{50-24}}{{\left( y \right)}^{50}}
On simplifying the exponent, we get:
⇒T25=50C24(3)26(y)50\Rightarrow {{T}_{25}}{{=}^{50}}{{C}_{24}}{{\left( \sqrt{3} \right)}^{26}}{{\left( y \right)}^{50}}
We will also compare it with the 26th{{26}^{th}} term to compare the value. On taking r=25r=25, we get:
⇒T25+1=50C25(3)50−25(y)50\Rightarrow {{T}_{25+1}}{{=}^{50}}{{C}_{25}}{{\left( \sqrt{3} \right)}^{50-25}}{{\left( y \right)}^{50}}
On simplifying, we get:
⇒T26=50C26(3)24(y)50\Rightarrow {{T}_{26}}{{=}^{50}}{{C}_{26}}{{\left( \sqrt{3} \right)}^{24}}{{\left( y \right)}^{50}}
On comparing the values of T25{{T}_{25}} and T26{{T}_{26}} we get the value of T25{{T}_{25}} higher therefore, in (x+y)50{{\left( x+y \right)}^{50}}the highest value is at T25{{T}_{25}} given ∣x∣=3∣y∣\left| x \right|=\sqrt{3}\left| y \right| therefore, the correct option is (D)\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.