Let $$\langle a_{n}\rangle$$ be a sequence such that $$a_{0}=0,a_{1}=\frac{1}{2}$$ and $$2a_{n+2}=5a_{n+1}-3a_{n},n=0,1,2,3,....$$ Then $$\sum_{k=1}^{100}a_{k}$$ is equal to

Step 1: Rearrange the recurrence relation

We are given:

$$2a_{n+2} = 5a_{n+1} - 3a_n$$

We can split the $$5a_{n+1}$$ term into $$2a_{n+1} + 3a_{n+1}$$ and rearrange the equation:

$$2a_{n+2} - 3a_{n+1} = 2a_{n+1} - 3a_n$$

Notice that the expression on the left is exactly the same as the expression on the right, just shifted to the next index. This means the sequence $$b_n = 2a_{n+1} - 3a_n$$ is a constant sequence.

Step 2: Find the constant value

Let's find the value of this constant using our initial conditions, $$a_0 = 0$$ and $$a_1 = \frac{1}{2}$$.

For $$n = 0$$:

$$2a_1 - 3a_0 = 2\left(\frac{1}{2}\right) - 3(0) = 1 - 0 = 1$$

Therefore, for any integer $$k \ge 1$$:

$$2a_k - 3a_{k-1} = 1$$

Step 3: Sum the relation

Now, take the sum of both sides of this new equation from $$k = 1$$ to $$100$$:

$$\sum_{k=1}^{100} (2a_k - 3a_{k-1}) = \sum_{k=1}^{100} 1$$

Split the sum on the left side:

$$2\sum_{k=1}^{100} a_k - 3\sum_{k=1}^{100} a_{k-1} = 100$$

Step 4: Express in terms of the total sum $$S$$

Let the sum we are trying to find be $$S = \sum_{k=1}^{100} a_k$$.

The second sum is $$\sum_{k=1}^{100} a_{k-1} = a_0 + a_1 + a_2 + \dots + a_{99}$$.

Notice that this is almost the entire sum $$S$$, except it includes $$a_0$$ and is missing $$a_{100}$$. Therefore, we can write it as:

$$\sum_{k=1}^{100} a_{k-1} = S - a_{100} + a_0$$

Substitute this back into our equation:

$$2S - 3(S - a_{100} + a_0) = 100$$

Step 5: Solve for $$S$$

Since we know $$a_0 = 0$$, substitute it in:

$$2S - 3(S - a_{100} + 0) = 100$$

$$2S - 3S + 3a_{100} = 100$$

$$-S + 3a_{100} = 100$$

Rearrange to isolate $$S$$:

$$S = 3a_{100} - 100$$

Therefore, the correct option is (B).