The problem asks us to find a closed formula for the sequence $(c_n): 0, 1, 3, 7, 15, 31, ...$. We are given two formulas to use: $T_n = \frac{n(n+1)}{2}$ and $a_n = 2^n$.

1. Problem Description

The problem asks us to find a closed formula for the sequence $(c_n): 0, 1, 3, 7, 15, 31, ...$. We are given two formulas to use: $T_n = \frac{n(n+1)}{2}$ and $a_n = 2^n$.

2. Solution Steps

Let's analyze the sequence. We have:

$c_0 = 0$

$c_1 = 1$

$c_2 = 3$

$c_3 = 7$

$c_4 = 15$

$c_5 = 31$

We observe that the terms in the sequence are close to powers of

2. Specifically:

$c_0 = 2^0 - 1 = 1 - 1 = 0$

$c_1 = 2^1 - 1 = 2 - 1 = 1$

$c_2 = 2^2 - 1 = 4 - 1 = 3$

$c_3 = 2^3 - 1 = 8 - 1 = 7$

$c_4 = 2^4 - 1 = 16 - 1 = 15$

$c_5 = 2^5 - 1 = 32 - 1 = 31$

Thus, we can deduce that the closed formula for the sequence is $c_n = 2^n - 1$.

Since the sequence starts with $c_0$, we will replace $n$ by $n$ in the formula $a_n = 2^n$. Then, $c_n = a_n - 1$.

Verification:

Let's verify the closed formula for the first few terms:

$c_0 = 2^0 - 1 = 1 - 1 = 0$

$c_1 = 2^1 - 1 = 2 - 1 = 1$

$c_2 = 2^2 - 1 = 4 - 1 = 3$

$c_3 = 2^3 - 1 = 8 - 1 = 7$

$c_4 = 2^4 - 1 = 16 - 1 = 15$

$c_5 = 2^5 - 1 = 32 - 1 = 31$

The values match the given sequence.

3. Final Answer

The closed formula for the sequence $(c_n): 0, 1, 3, 7, 15, 31, ...$ is $c_n = 2^n - 1$.