The total number of terms in the expansion of `(x+2)^(102) + (x-2)^(102)`, if similar terms are taken together are

Options

A. 51
B. 53
C. 54
D. 52

Answer

To find the total number of terms in the expansion of \((x+2)^{102} + (x-2)^{102}\) when similar terms are taken together, we will follow these steps:

### Step 1: Understand the expansions
The expression consists of two parts: \((x+2)^{102}\) and \((x-2)^{102}\). We will first expand each part using the Binomial Theorem.

### Step 2: Apply the Binomial Theorem
The Binomial Theorem states that:
\[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]
For \((x+2)^{102}\):
\[ (x+2)^{102} = \sum_{r=0}^{102} \binom{102}{r} x^{102-r} 2^r \]
For \((x-2)^{102}\):
\[ (x-2)^{102} = \sum_{r=0}^{102} \binom{102}{r} x^{102-r} (-2)^r \]

### Step 3: Combine the expansions
Now, we will add the two expansions:
\[ (x+2)^{102} + (x-2)^{102} = \sum_{r=0}^{102} \binom{102}{r} x^{102-r} (2^r + (-2)^r) \]
Notice that \(2^r + (-2)^r\) will yield:
- \(0\) if \(r\) is odd (since \(2^r - 2^r = 0\))
- \(2 \cdot 2^r = 2^{r+1}\) if \(r\) is even

### Step 4: Identify the terms that remain
Thus, only the even \(r\) terms will contribute to the final expression. The even values of \(r\) can be expressed as \(r = 2k\) where \(k\) can take values from \(0\) to \(51\) (since \(r\) can go up to \(102\)).

### Step 5: Count the remaining terms
The values of \(k\) range from \(0\) to \(51\), which gives us \(52\) possible values (including \(0\)). Therefore, there will be \(52\) terms in the combined expansion.

### Conclusion
The total number of terms in the expansion of \((x+2)^{102} + (x-2)^{102}\), when similar terms are taken together, is **52**.

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