We are given that $S_n$ is the sum of the first $n$ terms of a sequence $\{a_n\}$. We are also given that $S_{n+1} = S_n + a_n + 1$ and $a_2 + a_6 = 10$. We are asked to find $S_7$.

1. Problem Description

We are given that $S_n$ is the sum of the first $n$ terms of a sequence $\{a_n\}$. We are also given that $S_{n+1} = S_n + a_n + 1$ and $a_2 + a_6 = 10$. We are asked to find $S_7$.

2. Solution Steps

First, we have the relation $S_{n+1} = S_n + a_n + 1$. We know that $S_{n+1} = S_n + a_{n+1}$.

Therefore, $S_n + a_{n+1} = S_n + a_n + 1$, which implies $a_{n+1} = a_n + 1$.

This means that the sequence $\{a_n\}$ is an arithmetic sequence with a common difference $d=1$.

We are given $a_2 + a_6 = 10$. We can express $a_2$ and $a_6$ in terms of $a_1$ and $d$.

We know that $a_n = a_1 + (n-1)d$. Thus,

$a_2 = a_1 + (2-1)d = a_1 + d = a_1 + 1$.

$a_6 = a_1 + (6-1)d = a_1 + 5d = a_1 + 5$.

Therefore, $a_2 + a_6 = (a_1 + 1) + (a_1 + 5) = 2a_1 + 6 = 10$.

Solving for $a_1$, we have $2a_1 = 4$, so $a_1 = 2$.

Then, $a_n = 2 + (n-1)(1) = n+1$.

We want to find $S_7$. We have $S_7 = \sum_{n=1}^7 a_n = \sum_{n=1}^7 (n+1)$.

$S_7 = (1+1) + (2+1) + (3+1) + (4+1) + (5+1) + (6+1) + (7+1) = 2 + 3 + 4 + 5 + 6 + 7 + 8$.

$S_7 = \sum_{n=1}^7 (n+1) = \sum_{n=2}^8 n = \sum_{n=1}^8 n - 1 = \frac{8(8+1)}{2} - 1 = \frac{8 \cdot 9}{2} - 1 = \frac{72}{2} - 1 = 36-1 = 35$.

Alternatively,

$S_7 = 2 + 3 + 4 + 5 + 6 + 7 + 8 = 35$.

$S_{n+1} = S_n + a_n + 1$

$a_{n+1} = a_n + 1$

$a_n = a_1 + (n-1)d$

$a_2 + a_6 = 10$

$a_2 = a_1 + d$

$a_6 = a_1 + 5d$

$a_2 + a_6 = 2a_1 + 6d = 10$

$a_1 = 2$

$a_n = a_1 + (n-1)d = 2 + (n-1)1 = n+1$

$S_7 = \sum_{n=1}^7 a_n = \sum_{n=1}^7 (n+1)$

$S_7 = \sum_{n=1}^7 n + \sum_{n=1}^7 1 = \frac{7(7+1)}{2} + 7 = \frac{7 \cdot 8}{2} + 7 = 28 + 7 = 35$

3. Final Answer

35