The number of terms in the expansion of $$(1+x)^{101}(1-x+x^2)^{100}$$ in powers of $$x$$ is:

We are given the expression $$(1+x)^{101}(1-x+x^2)^{100}$$ and need to find the number of distinct powers of $$x$$ in its expansion.

First, recognize that $$1 - x + x^2$$ can be rewritten using the identity $$1 + x^3 = (1 + x)(1 - x + x^2)$$. Solving for $$1 - x + x^2$$, we get:

$$1 - x + x^2 = \frac{1 + x^3}{1 + x}$$

for $$x \neq -1$$. Substituting this into the original expression:

$$(1+x)^{101} \cdot (1-x+x^2)^{100} = (1+x)^{101} \cdot \left( \frac{1 + x^3}{1 + x} \right)^{100} = (1+x)^{101} \cdot \frac{(1 + x^3)^{100}}{(1 + x)^{100}}$$

Simplify the exponents:

$$= (1+x)^{101-100} \cdot (1 + x^3)^{100} = (1+x) \cdot (1 + x^3)^{100}$$

So, the expression simplifies to $$(1+x)(1 + x^3)^{100}$$.

Now, expand $$(1 + x^3)^{100}$$ using the binomial theorem:

$$(1 + x^3)^{100} = \sum_{k=0}^{100} \binom{100}{k} (x^3)^k = \sum_{k=0}^{100} \binom{100}{k} x^{3k}$$

Multiply this by $$(1 + x)$$:

$$(1+x) \cdot \sum_{k=0}^{100} \binom{100}{k} x^{3k} = \sum_{k=0}^{100} \binom{100}{k} x^{3k} + x \cdot \sum_{k=0}^{100} \binom{100}{k} x^{3k} = \sum_{k=0}^{100} \binom{100}{k} x^{3k} + \sum_{k=0}^{100} \binom{100}{k} x^{3k+1}$$

This gives two separate sums:

1. The first sum: $$\sum_{k=0}^{100} \binom{100}{k} x^{3k}$$ has exponents $$3k$$ for $$k = 0, 1, 2, \ldots, 100$$. This produces exponents: $$0, 3, 6, \ldots, 300$$.
2. The second sum: $$\sum_{k=0}^{100} \binom{100}{k} x^{3k+1}$$ has exponents $$3k+1$$ for $$k = 0, 1, 2, \ldots, 100$$. This produces exponents: $$1, 4, 7, \ldots, 301$$.

Now, count the number of terms in each sum:

• For the first sum, exponents form an arithmetic sequence with first term 0, common difference 3, and last term 300. The number of terms is $$\frac{300 - 0}{3} + 1 = 100 + 1 = 101$$.
• For the second sum, exponents form an arithmetic sequence with first term 1, common difference 3, and last term 301. The number of terms is $$\frac{301 - 1}{3} + 1 = \frac{300}{3} + 1 = 100 + 1 = 101$$.

To find the total number of distinct powers, check if there is any overlap between the exponents of the two sums. Suppose an exponent $$m$$ appears in both sums. Then:

$$m = 3a \quad$$ for some integer $$a \quad$$ (from the first sum)

$$m = 3b + 1 \quad$$ for some integer $$b \quad$$ (from the second sum)

Equating these:

$$3a = 3b + 1 \implies 3a - 3b = 1 \implies 3(a - b) = 1$$

This equation has no integer solutions because 3 times an integer cannot equal 1. Therefore, there are no common exponents between the two sums.

Since there is no overlap, the total number of distinct exponents is the sum of the terms from both sums:

$$101 + 101 = 202$$

Hence, the number of distinct powers of $$x$$ is 202.

So, the answer is 202.