Q.The total number of terms in the expansion of $(x + a)^{100} + (x - a)^{100}$ after simplification is
(A) $50$
(B) $202$
(C) $51$
(D) none of these

The key idea is that in the expansion of $(x + a)^{100} + (x - a)^{100}$, odd powers of $a$ cancel out, leaving only even powers.

Step 1: Write the general term of $(x + a)^{100}$ as $\binom{100}{r} x^{100-r} a^r$, and of $(x - a)^{100}$ as $\binom{100}{r} x^{100-r} (-a)^r = \binom{100}{r} x^{100-r} (-1)^r a^r$.

Step 2: Adding them, the terms where $r$ is odd cancel (since $1 + (-1)^r = 0$), and where $r$ is even they double: $2 \binom{100}{r} x^{100-r} a^r$. …

The Binomial Theorem tells us that $(x + a)^{100}$ expands into a sum of terms of the form $\binom{100}{r} x^{100-r} a^r$, where $r$ runs from $0$ to $100$. Similarly, $(x - a)^{100}$ expands into $\binom{100}{r} x^{100-r} (-a)^r = \binom{100}{r} x^{100-r} (-1)^r a^r$.

When we add these two expansions, something beautiful happens: terms where $r$ is odd have opposite signs and cancel out, while terms where $r$ is even have the same sign and add up. This is the core insight — the sum only retains the even-indexed terms.

Let’s work through it step by step.

1. Write the general term for each expansion.

   For $(x + a)^{100}$, the $r$-th term is $T_r = \binom{100}{r} x^{100-r} a^r$.

   For $(x - a)^{100}$, the $r$-th term is $U_r = \binom{100}{r} x^{100-r} (-a)^r = \binom{100}{r} x^{100-r} (-1)^r a^r$.

2. Add the corresponding terms.

   The sum of the two expansions is:

Combine them term by term:

3. Identify which terms survive. The factor $1 + (-1)^r$ is $2$ when $r$ is even, and $0$ when $r$ is odd. So only even $r$ contribute. This means the sum simplifies to:

4. Count the number of distinct terms.

   The even values of $r$ from $0$ to $100$ are $0, 2, 4, \dots, 100$. How many are there?

   This is an arithmetic progression: first term $0$, last term $100$, common difference $2$.

   Number of terms = $\frac{100 - 0}{2} + 1 = 50 + 1 = 51$.

5. Check for any further simplification. …