Alright class, let's dive deep into a very important concept within the Binomial Theorem: the
Middle Term(s) of an expansion. This isn't just about memorizing a formula; it's about understanding the structure of binomial expansions and how the terms are distributed. This concept frequently appears in JEE Main and Advanced, often subtly, as a part of a larger problem.
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Understanding the Middle Term(s) in Binomial Expansion
Recall the general expansion of a binomial expression $(a+b)^n$:
$$ (a+b)^n = inom{n}{0}a^n b^0 + inom{n}{1}a^{n-1} b^1 + inom{n}{2}a^{n-2} b^2 + dots + inom{n}{r}a^{n-r} b^r + dots + inom{n}{n}a^0 b^n $$
The general term, also known as the $(r+1)^{th}$ term, is given by $T_{r+1} = inom{n}{r} a^{n-r} b^r$.
An important characteristic of this expansion is that it contains a total of
$(n+1)$ terms. The concept of "middle term" arises from finding the term(s) that are centrally located within this sequence of $(n+1)$ terms. The coefficients, $inom{n}{r}$, are symmetric around the middle. For example, $inom{n}{0} = inom{n}{n}$, $inom{n}{1} = inom{n}{n-1}$, and so on. This symmetry makes the middle term(s) special.
The method to find the middle term(s) depends entirely on whether the exponent 'n' is an
even number or an
odd number. Let's explore both cases thoroughly.
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Case 1: When 'n' is an Even Integer
If the exponent 'n' is an even number, then the total number of terms in the expansion, which is $(n+1)$, will be an odd number.
Think about it: if n=4, total terms = 5 (an odd number).
When you have an odd number of items, there's always a single, distinct middle item.
Derivation of Position:
Let the total number of terms be $N = n+1$. Since $n$ is even, $N$ is odd.
For an odd number of terms, say $k$, the middle term's position is given by $frac{k+1}{2}$.
In our case, $k = n+1$. So, the position of the middle term is:
$$ ext{Position of Middle Term} = frac{(n+1)+1}{2} = frac{n+2}{2} = left(frac{n}{2} + 1
ight)^{th} ext{ term} $$
This means that if we are looking for the $(r+1)^{th}$ term, then $r+1 = frac{n}{2} + 1$.
Therefore, $r = frac{n}{2}$.
Formula for the Middle Term:
Substituting $r = frac{n}{2}$ into the general term formula $T_{r+1} = inom{n}{r} a^{n-r} b^r$, we get:
$$ extbf{Middle Term} = T_{n/2 + 1} = inom{n}{n/2} a^{n - n/2} b^{n/2} $$
$$ oxed{ extbf{Middle Term} = inom{n}{n/2} a^{n/2} b^{n/2}} $$
This is the only middle term when 'n' is even. Its binomial coefficient, $inom{n}{n/2}$, is the largest among all binomial coefficients in the expansion.
Example 1: Find the middle term in the expansion of $(x + 2y)^6$.
Here, $n=6$, which is an even number.
The total number of terms is $n+1 = 6+1 = 7$.
The position of the middle term is $left(frac{6}{2} + 1
ight)^{th} = (3+1)^{th} = 4^{th}$ term.
So, we need to find $T_4$. Using the general term $T_{r+1} = inom{n}{r} a^{n-r} b^r$, we set $r=3$, $n=6$, $a=x$, $b=2y$.
$$ T_4 = inom{6}{3} x^{6-3} (2y)^3 $$
$$ T_4 = inom{6}{3} x^3 (2^3 y^3) $$
$$ T_4 = frac{6 imes 5 imes 4}{3 imes 2 imes 1} x^3 (8y^3) $$
$$ T_4 = 20 imes 8 x^3 y^3 $$
$$ oxed{T_4 = 160 x^3 y^3} $$
Thus, $160 x^3 y^3$ is the middle term.
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Case 2: When 'n' is an Odd Integer
If the exponent 'n' is an odd number, then the total number of terms in the expansion, $(n+1)$, will be an even number.
Think about it: if n=5, total terms = 6 (an even number).
When you have an even number of items, there isn't a single 'middle' item; instead, there are two middle items.
Derivation of Positions:
Let the total number of terms be $N = n+1$. Since $n$ is odd, $N$ is even.
For an even number of terms, say $k$, the two middle terms are at positions $frac{k}{2}$ and $left(frac{k}{2} + 1
ight)$.
In our case, $k = n+1$. So, the positions of the two middle terms are:
1. First Middle Term's Position: $frac{n+1}{2}^{th}$ term
2. Second Middle Term's Position: $left(frac{n+1}{2} + 1
ight)^{th}$ term, which simplifies to $left(frac{n+3}{2}
ight)^{th}$ term.
Now, let's find the 'r' values for these terms using $T_{r+1}$.
For the first middle term, $r+1 = frac{n+1}{2}$, so $r = frac{n+1}{2} - 1 = frac{n+1-2}{2} = frac{n-1}{2}$.
For the second middle term, $r+1 = frac{n+3}{2}$, so $r = frac{n+3}{2} - 1 = frac{n+3-2}{2} = frac{n+1}{2}$.
Formulas for the Middle Terms:
Using $T_{r+1} = inom{n}{r} a^{n-r} b^r$:
1.
First Middle Term: (at position $frac{n+1}{2}$)
Substitute $r = frac{n-1}{2}$:
$$ extbf{First Middle Term} = T_{(n-1)/2 + 1} = inom{n}{(n-1)/2} a^{n - (n-1)/2} b^{(n-1)/2} $$
$$ oxed{ extbf{First Middle Term} = inom{n}{(n-1)/2} a^{(n+1)/2} b^{(n-1)/2}} $$
2.
Second Middle Term: (at position $frac{n+3}{2}$)
Substitute $r = frac{n+1}{2}$:
$$ extbf{Second Middle Term} = T_{(n+1)/2 + 1} = inom{n}{(n+1)/2} a^{n - (n+1)/2} b^{(n+1)/2} $$
$$ oxed{ extbf{Second Middle Term} = inom{n}{(n+1)/2} a^{(n-1)/2} b^{(n+1)/2}} $$
Notice that the binomial coefficients for these two terms are equal: $inom{n}{(n-1)/2} = inom{n}{n - (n-1)/2} = inom{n}{(n+1)/2}$. This is due to the property $inom{n}{k} = inom{n}{n-k}$.
Example 2: Find the middle terms in the expansion of $(x/3 + 9y)^5$.
Here, $n=5$, which is an odd number.
The total number of terms is $n+1 = 5+1 = 6$.
The positions of the two middle terms are $left(frac{5+1}{2}
ight)^{th} = 3^{rd}$ term and $left(frac{5+3}{2}
ight)^{th} = 4^{th}$ term.
So, we need to find $T_3$ and $T_4$. Here $a = x/3$ and $b = 9y$.
For the 3rd term ($T_3$, so $r=2$):
$$ T_3 = inom{5}{2} left(frac{x}{3}
ight)^{5-2} (9y)^2 $$
$$ T_3 = inom{5}{2} left(frac{x}{3}
ight)^3 (9y)^2 $$
$$ T_3 = 10 imes frac{x^3}{27} imes 81y^2 $$
$$ T_3 = 10 imes x^3 imes 3y^2 $$
$$ oxed{T_3 = 30 x^3 y^2} $$
For the 4th term ($T_4$, so $r=3$):
$$ T_4 = inom{5}{3} left(frac{x}{3}
ight)^{5-3} (9y)^3 $$
$$ T_4 = inom{5}{3} left(frac{x}{3}
ight)^2 (9y)^3 $$
$$ T_4 = 10 imes frac{x^2}{9} imes 729y^3 $$
$$ T_4 = 10 imes x^2 imes 81y^3 $$
$$ oxed{T_4 = 810 x^2 y^3} $$
Thus, $30 x^3 y^2$ and $810 x^2 y^3$ are the two middle terms.
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JEE Focus: Advanced Applications and Common Traps
While the basic identification of middle terms is straightforward, JEE problems often involve these concepts within more complex scenarios:
1.
Expressions with Negative Signs or Fractional Powers: Be very careful with signs and exponents.
Example 3: Find the middle term in the expansion of $left(frac{x}{a} - frac{a}{x}
ight)^{10}$.
Here $n=10$ (even). Total terms = 11.
Position of middle term: $left(frac{10}{2} + 1
ight)^{th} = 6^{th}$ term.
So, $r=5$. Here $A = x/a$ and $B = -a/x$.
$$ T_6 = inom{10}{5} left(frac{x}{a}
ight)^{10-5} left(-frac{a}{x}
ight)^5 $$
$$ T_6 = inom{10}{5} left(frac{x}{a}
ight)^5 left(-frac{a}{x}
ight)^5 $$
$$ T_6 = inom{10}{5} left(frac{x^5}{a^5}
ight) left(-frac{a^5}{x^5}
ight) $$
$$ T_6 = inom{10}{5} (-1) left(frac{x^5}{a^5} imes frac{a^5}{x^5}
ight) $$
$$ T_6 = -inom{10}{5} imes 1 $$
$$ inom{10}{5} = frac{10 imes 9 imes 8 imes 7 imes 6}{5 imes 4 imes 3 imes 2 imes 1} = 2 imes 9 imes 2 imes 7 = 252 $$
$$ oxed{T_6 = -252} $$
Notice how the variable terms cancelled out, resulting in a constant middle term. This is a common pattern for expansions of the form $(X + 1/X)^n$ or $(X - 1/X)^n$ when $n$ is even.
2.
Finding the Coefficient of the Middle Term: Sometimes only the numerical coefficient is asked. Make sure to identify $a$ and $b$ correctly and simplify.
Example 4: Find the coefficient of the middle term(s) in $(1-3x)^{13}$.
Here $n=13$ (odd). Total terms = 14.
The two middle terms are at positions $left(frac{13+1}{2}
ight)^{th} = 7^{th}$ and $left(frac{13+3}{2}
ight)^{th} = 8^{th}$.
So we need $T_7$ (where $r=6$) and $T_8$ (where $r=7$).
Here $a=1$ and $b=-3x$.
For the 7th term ($T_7$, $r=6$):
$$ T_7 = inom{13}{6} (1)^{13-6} (-3x)^6 $$
$$ T_7 = inom{13}{6} (1)^7 (-3)^6 x^6 $$
$$ T_7 = inom{13}{6} (729) x^6 $$
The coefficient is $inom{13}{6} imes 729$.
$$ inom{13}{6} = frac{13 imes 12 imes 11 imes 10 imes 9 imes 8}{6 imes 5 imes 4 imes 3 imes 2 imes 1} = 13 imes 11 imes 2 imes 3 imes frac{8}{6 imes 5 imes 4} imes frac{10}{5 imes 2} imes frac{12}{6 imes 4 imes 3} = 13 imes 11 imes 2 imes 3 imes frac{2}{3} imes frac{1}{1} imes frac{1}{1} imes 1 = 1716 $$
$$ ext{Coefficient of } T_7 = 1716 imes 729 $$
For the 8th term ($T_8$, $r=7$):
$$ T_8 = inom{13}{7} (1)^{13-7} (-3x)^7 $$
$$ T_8 = inom{13}{7} (1)^6 (-3)^7 x^7 $$
$$ T_8 = inom{13}{7} (-2187) x^7 $$
Since $inom{13}{7} = inom{13}{13-7} = inom{13}{6} = 1716$.
The coefficient is $inom{13}{7} imes (-2187) = 1716 imes (-2187)$.
Important Note: While $inom{13}{6}$ and $inom{13}{7}$ are equal, the powers of $(-3)$ are different ($6$ and $7$), leading to different numerical coefficients (one positive, one negative). So, the coefficients of the middle terms are $1716 imes 729$ and $1716 imes (-2187)$.
3.
Middle Term as the Greatest Term: For $(1+x)^n$ type expansions, the middle term(s) often have the greatest coefficient. However, the greatest *value* of the term depends on 'x' as well. For $(a+b)^n$, the greatest binomial coefficient is $inom{n}{n/2}$ (n even) or $inom{n}{(n-1)/2}, inom{n}{(n+1)/2}$ (n odd). Whether the term itself is greatest depends on the specific values of 'a' and 'b'. This is a separate, but related, concept often taught alongside middle terms.
4.
Contextual Problems: Often, finding the middle term is just one step in a multi-part problem, e.g., "If the middle term of $(x^2 + 1/x)^n$ is independent of x, find n." or "If the coefficient of the middle term of $(1+x)^{2n}$ is P, and the coefficient of the middle term of $(1+x)^{2n-1}$ are Q and R, show P = Q+R." Such problems require a solid grasp of the middle term formulas.
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Summary Table: Middle Term(s)
To consolidate your understanding, here's a quick reference table:
Condition on 'n' |
Total Number of Terms (N) |
Number of Middle Terms |
Position(s) of Middle Term(s) |
Form of Middle Term(s) $T_{r+1}$ |
|---|
n is even |
$n+1$ (odd) |
One |
$left(frac{n}{2} + 1
ight)^{th}$ |
$T_{n/2 + 1} = inom{n}{n/2} a^{n/2} b^{n/2}$ |
n is odd |
$n+1$ (even) |
Two |
$left(frac{n+1}{2}
ight)^{th}$ and $left(frac{n+3}{2}
ight)^{th}$ |
$T_{(n+1)/2} = inom{n}{(n-1)/2} a^{(n+1)/2} b^{(n-1)/2}$
$T_{(n+3)/2} = inom{n}{(n+1)/2} a^{(n-1)/2} b^{(n+1)/2}$ |
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Concluding Thoughts
The concept of middle terms might seem simple at first glance, but its applications in JEE are quite diverse. Always remember to:
1.
Identify 'n': Is it even or odd? This is the crucial first step.
2.
Determine the position(s): Use the correct formula for even/odd 'n'.
3.
Identify 'a' and 'b': Pay attention to signs and complex terms (like $x^2$ or $1/x$).
4.
Calculate the term: Use the general term formula $T_{r+1} = inom{n}{r} a^{n-r} b^r$ carefully, simplifying all powers and coefficients.
5.
Re-read the question: Are you asked for the term itself, its coefficient, or something else related to it?
Practice these problems diligently, and you'll find this topic to be a score booster in your exams!