# Exercise 7.2.1

Use the substitution method to prove that the recurrence $T(n) = T(n-1) + \Theta(n)$ has the solution $T(n) = \Theta(n^2)$, as claimed at the beginning of section 7.2

We represent $\Theta(n)$ as $c_2n$ and we guess that $T(n) \le c_1n^2$

\begin{aligned} T(n) &= T(n-1) + c_2n \\ &\le c_1(n-1)^2 + c_2n \\ &= c_1n^2 - 2c_1n + c_1 + c_2n & (2c_1 > c_2, n \ge c_1/(2c_1 - c_2))\\ &\le c_1n^2 \end{aligned}