Let $\langle a_n \rangle$ be a sequence such that $a_0 = 0$, $a_1 = \frac{1}{2}$ and $2 a_{n+2} = 5 a_{n+1} - 3 a_n$, $n = 0, 1, 2, 3, \ldots$. Then $\sum_{k=1}^{100} a_k$ is equal to

Step 1: Find the characteristic equation

The given recurrence relation is a linear homogeneous recurrence relation with constant coefficients. To find the general form of $a_n$, we first write its characteristic equation by replacing $a_{n+2}$ with $r^2$, $a_{n+1}$ with $r$, and $a_n$ with $1$.

Step 2: Solve the characteristic equation for roots

We solve the quadratic characteristic equation to find its roots. Factoring the quadratic equation $2r^2 - 5r + 3 = 0$ gives us $(2r - 3)(r - 1) = 0$. This yields two distinct roots, $r_1 = \frac{3}{2}$ and $r_2 = 1$.

Step 3: Determine the general form of $a_n$

Since the roots are distinct, the general form of the sequence $a_n$ is $A r_1^n + B r_2^n$, where $A$ and $B$ are constants. Substituting the roots, we get $a_n = A \left(\frac{3}{2}\right)^n + B (1)^n$.

Step 4: Use initial conditions to find A and B

We use the given initial conditions $a_0 = 0$ and $a_1 = \frac{1}{2}$ to find the values of $A$ and $B$. For $n=0$: $a_0 = A(3/2)^0 + B(1)^0 = A + B = 0 \implies B = -A$. For $n=1$: $a_1 = A(3/2)^1 + B(1)^1 = \frac{3}{2}A + B = \frac{1}{2}$. Substituting $B = -A$ into the second equation: $\frac{3}{2}A - A = \frac{1}{2} \implies \frac{1}{2}A = \frac{1}{2} \implies A = 1$. Then $B = -1$. Thus, $a_n = \left(\frac{3}{2}\right)^n - 1$.

Step 5: Calculate the sum of the sequence

We need to find the sum of the first 100 terms of the sequence, starting from $k=1$. We can split the sum into two parts: a geometric series and a sum of constants. $$\sum_{k=1}^{100} \left(\frac{3}{2}\right)^k - \sum_{k=1}^{100} 1$$ The first part is a geometric series with first term $a = \frac{3}{2}$, common ratio $r = \frac{3}{2}$, and $n = 100$ terms. The sum of a geometric series is $S_n = a \frac{r^n - 1}{r - 1}$. So, $S_{100} = \frac{3}{2} \frac{\left(\frac{3}{2}\right)^{100} - 1}{\frac{3}{2} - 1} = \frac{3}{2} \frac{\left(\frac{3}{2}\right)^{100} - 1}{\frac{1}{2}} = 3 \left[\left(\frac{3}{2}\right)^{100} - 1\right]$. The second part is $\sum_{k=1}^{100} 1 = 100$. Therefore, the total sum is $3 \left[\left(\frac{3}{2}\right)^{100} - 1\right] - 100$.

Step 6: Simplify the sum

Simplifying the expression obtained in the previous step, we distribute the 3 and combine the constant terms.

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