The sum of coefficients of the two middle terms in the expansion of $(1 + x)^{2n - 1}$ is equal to ${}^{2n - 1}C_n$.

> [!TLDR] > The expansion of $(1+x)^{2n-1}$ has $2n$ terms, with the $n$-th and $(n+1)$-th terms being the middle terms. Their coefficients are ${}^{2n-1}C_{n-1}$ and ${}^{2n-1}C_n$ respectively. Using Pascal's Identity, their sum is ${}^{2n}C_n$. Since ${}^{2n}C_n \neq {}^{2n-1}C_n$, the given statement is **false**. The problem asks us to verify if a specific statement about the binomial expansion of $(1+x)^{2n-1}$ is true. The statement claims that the sum of the coefficients of its two middle terms is equal to ${}^{2n-1}C_n$. To determine this, we need to understand how to find the middle terms in a binomial expansion and how to calculate their coefficients. When we expand a binomial $(a+b)^N$, there are $N+1$ terms in total. The general term, often denoted as $T_{r+1}$, is given by ${}^N C_r a^{N-r} b^r$. The coefficient of this term is ${}^N C_r$. The number of middle terms depends on whether the exponent $N$ is even or odd: * If $N$ is even, there is one middle term at position $\frac{N}{2} + 1$. * If $N$ is odd, there are two middle terms at positions $\frac{N+1}{2}$ and $\frac{N+1}{2} + 1$. In our problem, the exponent is $N = 2n-1$. This is an odd number. Therefore, there will be two middle terms. Let's proceed step-by-step to find these terms and their coefficients. 1. **Determine the total number of terms:** For the expansion of $(1+x)^{2n-1}$, the exponent is $N = 2n-1$. The total number of terms in the expansion is $N+1 = (2n-1)+1 = 2n$. 2. **Identify the positions of the middle terms:** Since the total number of terms ($2n$) is an even number, there are two middle terms. Their positions are $\frac{2n}{2} = n$ and $\frac{2n}{2} + 1 = n+1$. So, the $n$-th term ($T_n$) and the $(n+1)$-th term ($T_{n+1}$) are the middle terms. 3. **Find the $n$-th term ($T_n$) and its coefficient:** The general term in the expansion of $(1+x)^N$ is $T_{r+1} = {}^N C_r (1)^{N-r} (x)^r = {}^N C_r x^r$. For the $n$-th term, we set $

The sum of coefficients of the two middle terms in – Class 11 Mathematics

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(1 + x)^{2n - 1}

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Concept understanding — Binomial Theorem Expansion

The Problem That Started It All

Imagine you have to expand $(x + y)^2$ . That's easy: $x^2 + 2xy + y^2$ . Now try $(x + y)^3$ : $x^3 + 3x^2y + 3xy^2 + y^3$ . Still manageable.

But what about $(x + y)^{10}$ ? Or $(x + y)^{100}$ ? Multiplying it out term by term would take forever. There has to be a pattern — and there is.

The Binomial Theorem is the shortcut that tells you exactly what each term in the expansion of $(x + y)^n$ looks like, without ever having to multiply.

The Pattern You Already Know

Look at the expansions you already know:

Power	Expansion
$(x + y)^0$	$1$
$(x + y)^1$	$x + y$
$(x + y)^2$	$x^2 + 2xy + y^2$
$(x + y)^3$	$x^3 + 3x^2y + 3xy^2 + y^3$
$(x + y)^4$	$x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$

Notice three things:

The powers of $x$ decrease from $n$ down to $0$ , while the powers of $y$ increase from $0$ up to $n$ . In every term, the exponents add to $n$ .
The coefficients — 1, 4, 6, 4, 1 for $n=4$ — follow a famous pattern called Pascal's triangle.
The number of terms is always $n+1$ .

Note

Pascal's triangle: each number is the sum of the two numbers directly above it.

       1
      1 1
     1 2 1
    1 3 3 1
   1 4 6 4 1
  1 5 10 10 5 1

The Precise Statement

n

Let's break that down.

The symbol $\sum_{k=0}^{n}$ means "add up terms for $k = 0, 1, 2, \dots, n$ ". For each $k$ , the term is:

$\binom{n}{k}$ — the coefficient (read as "n choose k")
$x^{n-k}$ — $x$ raised to the power $n-k$
$y^k$ — $y$ raised to the power $k$

So for $n=4$ , the terms are:

$k$	$\binom{4}{k}$	$x^{4-k}$	$y^k$	Term
0	$\binom{4}{0}=1$	$x^4$	$y^0$	$1 \cdot x^4$
1	$\binom{4}{1}=4$	$x^3$	$y^1$	$4x^3y$
2	$\binom{4}{2}=6$	$x^2$	$y^2$	$6x^2y^2$
3	$\binom{4}{3}=4$	$x^1$	$y^3$	$4xy^3$
4	$\binom{4}{4}=1$	$x^0$	$y^4$	$1 \cdot y^4$

Add them up: $x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$ . Exactly what we had.

Where Do Those Coefficients Come From?

The binomial coefficient $\binom{n}{k}$ counts how many ways you can choose $k$ items from a set of $n$ items. In the expansion, it counts how many ways you can pick $k$ copies of $y$ (and therefore $n-k$ copies of $x$ ) when multiplying $(x+y)$ by itself $n$ times.

Tip

To compute $\binom{n}{k}$ quickly: start at $n$ and multiply $k$ decreasing numbers, then divide by $k!$ .

Example: $\binom{7}{3} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = 35$ .

The General Term

The $k$ -th term (starting with $k=0$ ) in the expansion is:

\text{General term: } T_{k+1} = \binom{n}{k} x^{n-k} y^k

This is the most useful part for exams. If someone asks "find the 5th term in $(x+y)^{10}$ ", you set $k=4$ (because $T_{k+1}$ means $k=4$ gives the 5th term) and write:

T_5 = \binom{10}{4} x^{6} y^{4}

What If It's Not Just $x$ and $y$ ? …

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The expansion of $(1+x)^{2n-1}$ has $2n$ terms, with the $n$ -th and $(n+1)$ -th terms being the middle terms. Their coefficients are ${}^{2n-1}C_{n-1}$ and ${}^{2n-1}C_n$ respectively. Using Pascal's Identity, their sum is ${}^{2n}C_n$ . Since ${}^{2n}C_n \neq {}^{2n-1}C_n$ , the given statement is false.

The problem asks us to verify if a specific statement about the binomial expansion of $(1+x)^{2n-1}$ is true. The statement claims that the sum of the coefficients of its two middle terms is equal to ${}^{2n-1}C_n$ . To determine this, we need to understand how to find the middle terms in a binomial expansion and how to calculate their coefficients.

When we expand a binomial $(a+b)^N$ , there are $N+1$ terms in total. The general term, often denoted as $T_{r+1}$ , is given by ${}^N C_r a^{N-r} b^r$ . The coefficient of this term is ${}^N C_r$ .

The number of middle terms depends on whether the exponent $N$ is even or odd:

If $N$ is even, there is one middle term at position $\frac{N}{2} + 1$ .
If $N$ is odd, there are two middle terms at positions $\frac{N+1}{2}$ and $\frac{N+1}{2} + 1$ .

In our problem, the exponent is $N = 2n-1$ . This is an odd number. Therefore, there will be two middle terms.

Let's proceed step-by-step to find these terms and their coefficients.

Determine the total number of terms:

For the expansion of $(1+x)^{2n-1}$ , the exponent is $N = 2n-1$ .

The total number of terms in the expansion is $N+1 = (2n-1)+1 = 2n$ .
Identify the positions of the middle terms:

Since the total number of terms ( $2n$ ) is an even number, there are two middle terms.

Their positions are $\frac{2n}{2} = n$ and $\frac{2n}{2} + 1 = n+1$ .

So, the $n$ -th term ( $T_n$ ) and the $(n+1)$ -th term ( $T_{n+1}$ ) are the middle terms.
Find the $n$ -th term ( $T_n$ ) and its coefficient:

The general term in the expansion of $(1+x)^N$ is $T_{r+1} = {}^N C_r (1)^{N-r} (x)^r = {}^N C_r x^r$ .

For the $n$ -th term, we set $r+1 = n$ , which means $r = n-1$ .

Substituting $N = 2n-1$ and $r = n-1$ :

$T_n = T_{(n-1)+1} = {}^{2n-1}C_{n-1} x^{n-1}$ .

The coefficient of the $n$ -th term is ${}^{2n-1}C_{n-1}$ .
Find the $(n+1)$ -th term ( $T_{n+1}$ ) and its coefficient:

For the $(n+1)$ -th term, we set $r+1 = n+1$ , which means $r = n$ .

Substituting $N = 2n-1$ and $r = n$ :

$T_{n+1} = T_{n+1} = {}^{2n-1}C_n x^n$ .

The coefficient of the $(n+1)$ -th term is ${}^{2n-1}C_n$ .
Calculate the sum of the coefficients of the two middle terms:

The sum of these coefficients is ${}^{2n-1}C_{n-1} + {}^{2n-1}C_n$ .
Apply Pascal's Identity:

We use the identity for binomial coefficients: …

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Q.The sum of coefficients of the two middle terms in the expansion of (1+x)2n−1(1 + x)^{2n - 1}(1+x)2n−1 is equal to 2n−1Cn{}^{2n - 1}C_n2n−1Cn​.

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Q.The sum of coefficients of the two middle terms in the expansion of $(1 + x)^{2n - 1}$ is equal to ${}^{2n - 1}C_n$ .

Unlock everything free for 14 days