Q.Number of terms in the expansion of $(a + b)^n$ where $n \in \mathbb{N}$ is one less than the power $n$.

The given statement is false. We will prove by induction that the number of terms in the expansion of $(a+b)^n$ is $n+1$.

Concept: Proof By Induction.

Step 1: Base Case ($n=1$)

For $n=1$, the expansion is $(a+b)^1 = a+b$. This expansion has $2$ terms.

Our formula $n+1$ gives $1+1=2$. The base case holds.

Step 2: Inductive Hypothesis

Assume that for some natural number $k \ge 1$, the expansion of $(a+b)^k$ has $k+1$ terms.

This means $(a+b)^k = \sum_{r=0}^k \binom{k}{r} a^{k-r} b^r$.

Step 3: Inductive Step

Consider the expansion of $(a+b)^{k+1}$:

$(a+b)^{k+1} = (a+b)^k \cdot (a+b)$

Using the inductive hypothesis:

$(a+b)^{k+1} = \left(\sum_{r=0}^k \binom{k}{r} a^{k-r} b^r\right) \cdot (a+b)$

$(a+b)^{k+1} = \sum_{r=0}^k \binom{k}{r} a^{k-r+1} b^r + \sum_{r=0}^k \binom{k}{r} a^{k-r} b^{r+1}$

By combining like terms and using Pascal's Identity $\binom{k}{r} + \binom{k}{r-1} = \binom{k+1}{r}$, this simplifies to: …

The problem asks us to evaluate the truth of the statement: "Number of terms in the expansion of $(a + b)^n$ where $n \in \mathbb{N}$ is one less than the power $n$." To do this, we need to understand how terms are formed in a binomial expansion and then count them accurately.

The core concept here is the Binomial Theorem, which provides a systematic way to expand expressions of the form $(a+b)^n$. Each term in the expansion is unique in its combination of powers of $a$ and $b$, meaning they cannot be combined further.

Let's break down the analysis:

1. Understanding the Claim:

   The statement asserts that the number of terms is $n-1$. This means if $n=1$, there would be $1-1=0$ terms. If $n=2$, there would be $2-1=1$ term. This immediately raises a red flag, as even simple expansions like $(a+b)^1$ clearly have terms.

2. Testing with Small Values of $n$ (Counterexamples):

   Let's check the statement for the first few natural numbers:

   • For $n=1$: The expansion of $(a+b)^1$ is $a+b$. This expansion has 2 terms ($a$ and $b$). According to the statement, it should have $1-1=0$ terms. This is false.
   • For $n=2$: The expansion of $(a+b)^2$ is $a^2 + 2ab + b^2$. This expansion has 3 terms ($a^2$, $2ab$, and $b^2$). According to the statement, it should have $2-1=1$ term. This is false.
   • For $n=3$: The expansion of $(a+b)^3$ is $a^3 + 3a^2b + 3ab^2 + b^3$. This expansion has 4 terms. According to the statement, it should have $3-1=2$ terms. This is false.

   These examples clearly show that the given statement is incorrect. In each case, the actual number of terms is $n+1$, not $n-1$.

3. Applying the Binomial Theorem:

   The Binomial Theorem provides the general formula for the expansion of $(a+b)^n$:

   $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

   where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is the binomial coefficient.

   Let's write out the terms explicitly by substituting values for $k$:

   • For $k=0$: $\binom{n}{0}a^{n-0}b^0 = \binom{n}{0}a^n$
   • For $k=1$: $\binom{n}{1}a^{n-1}b^1$
   • For $k=2$: $\binom{n}{2}a^{n-2}b^2$
   • ...
   • For $k=n-1$: $\binom{n}{n-1}a^{n-(n-1)}b^{n-1} = \binom{n}{n-1}a^1 b^{n-1}$
   • For $k=n$: $\binom{n}{n}a^{n-n}b^n = \binom{n}{n}b^n$

   The full expansion is:

   $(a+b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \dots + \binom{n}{n-1}ab^{n-1} + \binom{n}{n}b^n$ …