Find the number of terms in the expansion of `(x + 2y)^20 + (x - 2y)^20`

Answer

To find the number of terms in the expansion of \((x + 2y)^{20} + (x - 2y)^{20}\), we will follow these steps:

### Step 1: Expand both expressions using the Binomial Theorem
The Binomial Theorem states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
Applying this to both expansions:

1. For \((x + 2y)^{20}\):
\[ (x + 2y)^{20} = \sum_{k=0}^{20} \binom{20}{k} x^{20-k} (2y)^k = \sum_{k=0}^{20} \binom{20}{k} x^{20-k} 2^k y^k \]

2. For \((x - 2y)^{20}\):
\[ (x - 2y)^{20} = \sum_{k=0}^{20} \binom{20}{k} x^{20-k} (-2y)^k = \sum_{k=0}^{20} \binom{20}{k} x^{20-k} (-2)^k y^k \]

### Step 2: Combine the two expansions
Now, we add the two expansions:
\[ (x + 2y)^{20} + (x - 2y)^{20} = \sum_{k=0}^{20} \binom{20}{k} x^{20-k} (2^k + (-2)^k) y^k \]

### Step 3: Analyze the terms
Notice that \(2^k + (-2)^k\) will yield:
- \(0\) for odd \(k\) (since \(2^k\) and \((-2)^k\) will cancel each other)
- \(2 \cdot 2^k\) for even \(k\) (since both terms will add up)

Thus, only the terms where \(k\) is even will contribute to the final sum.

### Step 4: Determine the values of \(k\)
The even values of \(k\) from \(0\) to \(20\) are:
\[ 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 \]
This gives us a total of \(11\) even terms.

### Conclusion
Therefore, the number of terms in the expansion of \((x + 2y)^{20} + (x - 2y)^{20}\) is \(11\).