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

We need to find $$\sum_{k=1}^{100} a_k$$ given $$a_0 = 0$$, $$a_1 = \frac{1}{2}$$, and $$2a_{n+2} = 5a_{n+1} - 3a_n$$.

Rearranging the recurrence relation yields $$2a_{n+2} - 5a_{n+1} + 3a_n = 0$$.

Substituting $$a_n = r^n$$ into this relation gives the characteristic equation $$2r^2 - 5r + 3 = 0$$ whose roots are $$r = \frac{5 \pm \sqrt{25 - 24}}{4} = \frac{5 \pm 1}{4}$$, i.e., $$r_1 = \frac{3}{2}$$ and $$r_2 = 1$$.

Hence the general solution is $$a_n = A\left(\frac{3}{2}\right)^n + B\cdot 1^n = A\left(\frac{3}{2}\right)^n + B$$.

Applying the initial condition $$a_0 = 0$$ gives $$A + B = 0 \implies B = -A$$, and substituting $$a_1 = \frac{1}{2}$$ yields $$\frac{3A}{2} + B = \frac{1}{2}$$. Substituting $$B = -A$$ into this equation gives $$\frac{3A}{2} - A = \frac{A}{2} = \frac{1}{2} \implies A = 1$$, hence $$B = -1$$. It follows that $$a_n = \left(\frac{3}{2}\right)^n - 1$$.

We then compute the sum as $$\sum_{k=1}^{100} a_k = \sum_{k=1}^{100}\left[\left(\frac{3}{2}\right)^k - 1\right] = \sum_{k=1}^{100}\left(\frac{3}{2}\right)^k - 100$$.

The geometric series evaluates to $$\sum_{k=1}^{100}\left(\frac{3}{2}\right)^k = \frac{3/2 \left[(3/2)^{100} - 1\right]}{3/2 - 1} = \frac{(3/2)\left[(3/2)^{100} - 1\right]}{1/2} = 3\left[\left(\frac{3}{2}\right)^{100} - 1\right]$$.

Substituting this result back gives $$\sum_{k=1}^{100} a_k = 3\left[\left(\frac{3}{2}\right)^{100} - 1\right] - 100 = 3\left(\frac{3}{2}\right)^{100} - 3 - 100 = 3\left(\frac{3}{2}\right)^{100} - 103$$.

Since $$a_{100} = \left(\frac{3}{2}\right)^{100} - 1$$, it follows that $$\left(\frac{3}{2}\right)^{100} = a_{100} + 1$$ and hence $$\sum_{k=1}^{100} a_k = 3(a_{100} + 1) - 103 = 3a_{100} + 3 - 103 = 3a_{100} - 100$$.

The correct answer is Option (2): $$3a_{100} - 100$$.