A sequence $x_1, x_2, x_3, \ldots,$ is defined by $x_1 = 1$ $x_{n+1} = (x_n)^2 - kx_n, \quad n \geq 1$ where $k$ is a constant, $k \neq 0$ - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 2

Using the expression found for $x_2$:

$x_2 = 1 - k,$

we substitute $x_2$ into the equation for $x_3$:

$x_3 = (x_2)^2 - kx_2$

Substituting $x_2 = 1 - k$:

$x_3 = (1 - k)^2 - k(1 - k)$

Expanding $(1 - k)^2$ gives:

$(1 - 2k + k^2) - (k - k^2) = 1 - 2k + k^2 - k + k^2$

Combining like terms:

$x_3 = 1 - 3k + 2k^2.$

Therefore, we have shown that:

$x_3 = 1 - 3k + 2k^2.$

Given that $x_3 = 1$, we set up the equation:

$1 - 3k + 2k^2 = 1.$

This simplifies to:

$2k^2 - 3k = 0.$

Factoring out $k$ gives:

$k(2k - 3) = 0.$

This means either $k = 0$ or $k = \frac{3}{2}$. Since $k \neq 0$, we find:

$k = \frac{3}{2}.$

With the value of $k$ found, we substitute into the expression for $x_n$:

The formula for the sequence becomes:

$x_n = 1 - 3 \left(\frac{3}{2}\right) + 2 \left(\frac{3}{2}\right)^2.$

Calculating this:

1. Compute $\left(\frac{3}{2}\right)^2 = \frac{9}{4}$.
2. Then, substituting:

$x_n = 1 - \frac{9}{2} + \frac{9}{2} = 1.$

Since all terms $x_n$ for $n\geq 1$ are equal to 1, we find:

$\sum_{n=1}^{100} x_n = 100.$