Given $B_{k+1}$ be obtained from $B_k$ using the symmetric rank-one update formula. Assume that the associated quasi-Newton method is applied to an n-dimensional, strictly convex, quadratic function, and that the vectors $s_{0}, . .., s_{n−1}$ are linearly independent, where $s_k = x_{k+1} - x_{k}$. Also, $(y_i − B_{i}s_{i})^{T}s_i = 0$ for all $i$. Show that the method terminates in at most $n + 1$ iterations.
My attempt: I was able to establish a relationship: $B_{k+1}s_i = y_i$ $\forall\ i=0,1,2,...,k$. Now, assume that the quasi-Newton method hasn't terminated after $n$ iterations (otherwise, we got what we need). I want to show that $\bigtriangledown f(x_{n+1}) = 0$, where $\bigtriangledown f(x) = Qx + c$ (since $f$ is a quadratic function). Now, by the notation of quasi-newton's method, since $y_n = \bigtriangledown f(x_{n+1}) - \bigtriangledown f(x_{n}) = Q(x_{n+1} - x_n)$, we have: $B_{n+1}s_n = y_n = Q(x_{n+1}-x_{n})$.
I got stucked on this problem over the last two days, and couldn't see how to proceed further. Can someone please help me with this problem? I would really appreciate any help.