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

2. Solution Steps

First, we need to identify a pattern in the given sequence.

The sequence is 0, 1, 3, 7, 15, 31, ... We can see that each term is one less than a power of

2. Specifically:

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

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

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

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

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

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

So, it seems like $c_n = 2^n - 1$, where $n$ starts from

0.

Let's verify this formula.

For n = 0, $c_0 = 2^0 - 1 = 1 - 1 = 0$.

For n = 1, $c_1 = 2^1 - 1 = 2 - 1 = 1$.

For n = 2, $c_2 = 2^2 - 1 = 4 - 1 = 3$.

For n = 3, $c_3 = 2^3 - 1 = 8 - 1 = 7$.

For n = 4, $c_4 = 2^4 - 1 = 16 - 1 = 15$.

For n = 5, $c_5 = 2^5 - 1 = 32 - 1 = 31$.

The closed formula $c_n = 2^n - 1$ works perfectly.

3. Final Answer

$c_n = 2^n - 1$