The coefficient of $x^n$ in the expansion of $(1 + x)^{2n}$ and $(1 + x)^{2n - 1}$ are in the ratio (A) $1 : 2$ (B) $1 : 3$ (C) $3 : 1$ (D) $2 : 1$

> [!TLDR] > The coefficient of $x^n$ in $(1+x)^{2n}$ is $\binom{2n}{n}$, and in $(1+x)^{2n-1}$ it is $\binom{2n-1}{n}$. Using the symmetry property $\binom{2n-1}{n} = \binom{2n-1}{n-1}$ and Pascal’s identity, the ratio simplifies to $2:1$, so the answer is **(D)**. The key insight here is the **symmetry property of combinations**: $\binom{n}{r} = \binom{n}{n-r}$. This, combined with Pascal’s identity, lets us relate coefficients across expansions without brute-force computation. --- 1. **Write the coefficients directly from the binomial theorem.** For $(1+x)^{2n}$, the general term is $\binom{2n}{r} x^r$. The coefficient of $x^n$ is $\binom{2n}{n}$. For $(1+x)^{2n-1}$, the general term is $\binom{2n-1}{r} x^r$. The coefficient of $x^n$ is $\binom{2n-1}{n}$. 2. **Apply symmetry to the second coefficient.** Since $\binom{2n-1}{n} = \binom{2n-1}{(2n-1)-n} = \binom{2n-1}{n-1}$, we can rewrite the ratio as: \[ \frac{\binom{2n}{n}}{\binom{2n-1}{n-1}} \] 3. **Use Pascal’s identity to connect the two.** Pascal’s rule says: $\binom{2n}{n} = \binom{2n-1}{n-1} + \binom{2n-1}{n}$. But from step 2, $\binom{2n-1}{n} = \binom{2n-1}{n-1}$, so: \[ \binom{2n}{n} = \binom{2n-1}{n-1} + \binom{2n-1}{n-1} = 2 \cdot \binom{2n-1}{n-1} \] 4. **Now the ratio is immediate.** \[ …

The coefficient of xⁿ in the expansion of (1 + x)² – Class 11 Mathematics

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Concept understanding — Combinations Symmetry Property

The Intuition: Two Ways to Choose

Imagine you have a group of 10 friends, and you need to pick 3 of them to form a committee. One way to think about this is: you are choosing the 3 people who will be on the committee. But there is another, equally valid way to think about it: you are rejecting the 7 people who will not be on the committee.

Choosing 3 to include is the same decision as choosing 7 to exclude. Every time you pick a set of 3, you automatically determine the set of 7 who are left out. There is a perfect one-to-one match between the two choices.

This is the heart of the symmetry property: the number of ways to choose $k$ items from $n$ is exactly the same as the number of ways to choose $n-k$ items from $n$ .

The Precise Statement

\binom{n}{k} = \binom{n}{n-k}

Where $\binom{n}{k}$ (read "n choose k") is the number of combinations — the number of distinct subsets of size $k$ you can pick from a set of $n$ distinct objects.

This holds for any non-negative integers $n$ and $k$ where $0 \leq k \leq n$ .

Why It Works (The Algebraic Proof)

The formula for combinations is:

\binom{n}{k} = \frac{n!}{k!(n-k)!}

Now compute $\binom{n}{n-k}$ :

\binom{n}{n-k} = \frac{n!}{(n-k)!(n-(n-k))!} = \frac{n!}{(n-k)!k!}

The denominator is just $k!(n-k)!$ written in a different order. Since multiplication is commutative, the two expressions are identical.

Note

The symmetry is purely algebraic, but the intuition is what makes it memorable: choosing $k$ to keep is the same as choosing $n-k$ to discard.

Special Cases That Make Sense

$k = 0$ : $\binom{n}{0} = 1$ (there is exactly one way to choose nothing). By symmetry, $\binom{n}{n} = 1$ (one way to choose everything). Both make sense — you either take nothing or take all.
$k = 1$ : $\binom{n}{1} = n$ . Symmetry gives $\binom{n}{n-1} = n$ . Choosing 1 person to include is the same as choosing $n-1$ people to exclude — there are $n$ choices in either case. …

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The coefficient of $x^n$ in $(1+x)^{2n}$ is $\binom{2n}{n}$ , and in $(1+x)^{2n-1}$ it is $\binom{2n-1}{n}$ . Using the symmetry property $\binom{2n-1}{n} = \binom{2n-1}{n-1}$ and Pascal’s identity, the ratio simplifies to $2:1$ , so the answer is (D).

The key insight here is the symmetry property of combinations: $\binom{n}{r} = \binom{n}{n-r}$ . This, combined with Pascal’s identity, lets us relate coefficients across expansions without brute-force computation.

Write the coefficients directly from the binomial theorem.

For $(1+x)^{2n}$ , the general term is $\binom{2n}{r} x^r$ . The coefficient of $x^n$ is $\binom{2n}{n}$ .

For $(1+x)^{2n-1}$ , the general term is $\binom{2n-1}{r} x^r$ . The coefficient of $x^n$ is $\binom{2n-1}{n}$ .
Apply symmetry to the second coefficient.

Since $\binom{2n-1}{n} = \binom{2n-1}{(2n-1)-n} = \binom{2n-1}{n-1}$ , we can rewrite the ratio as:

\frac{\binom{2n}{n}}{\binom{2n-1}{n-1}}

Use Pascal’s identity to connect the two. Pascal’s rule says: $\binom{2n}{n} = \binom{2n-1}{n-1} + \binom{2n-1}{n}$ . But from step 2, $\binom{2n-1}{n} = \binom{2n-1}{n-1}$ , so:

\binom{2n}{n} = \binom{2n-1}{n-1} + \binom{2n-1}{n-1} = 2 \cdot \binom{2n-1}{n-1}

Now the ratio is immediate.

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Q.The coefficient of xnx^nxn in the expansion of (1+x)2n(1 + x)^{2n}(1+x)2n and (1+x)2n−1(1 + x)^{2n - 1}(1+x)2n−1 are in the ratio (A) 1:21 : 21:2 (B) 1:31 : 31:3 (C) 3:13 : 13:1 (D) 2:12 : 12:1

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Q.The coefficient of $x^n$ in the expansion of $(1 + x)^{2n}$ and $(1 + x)^{2n - 1}$ are in the ratio
(A) $1 : 2$
(B) $1 : 3$
(C) $3 : 1$
(D) $2 : 1$

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