To solve for $s_7$, we need to evaluate the expression given by:

$s_k = \sum_{n=1}^{k} n (0.1)^n$

where $k = 7$. This means we need to compute:

$s_7 = 1(0.1)^1 + 2(0.1)^2 + 3(0.1)^3 + 4(0.1)^4 + 5(0.1)^5 + 6(0.1)^6 + 7(0.1)^7$

Let's calculate each term:

1. $1 \cdot (0.1)^1 = 1 \cdot 0.1 = 0.1$
2. $2 \cdot (0.1)^2 = 2 \cdot 0.01 = 0.02$
3. $3 \cdot (0.1)^3 = 3 \cdot 0.001 = 0.003$
4. $4 \cdot (0.1)^4 = 4 \cdot 0.0001 = 0.0004$
5. $5 \cdot (0.1)^5 = 5 \cdot 0.00001 = 0.00005$
6. $6 \cdot (0.1)^6 = 6 \cdot 0.000001 = 0.000006$
7. $7 \cdot (0.1)^7 = 7 \cdot 0.0000001 = 0.0000007$

Adding these together gives:

$s_7 = 0.1 + 0.02 + 0.003 + 0.0004 + 0.00005 + 0.000006 + 0.0000007$

Calculating the sum:

$s_7 = 0.1234567$

Thus, $s_7 = 0.1234567$.

Would you like further details or have any questions? Here are some related questions for practice:

1. How would you calculate $s_8$ using the same formula?
2. What is the general approach to summing a series involving powers and coefficients?
3. How does the value of $s_k$ behave as $k$ increases?
4. Can you generalize the formula for $s_k$ for an arbitrary value of $k$?
5. How would the calculation change if the base was $0.2$ instead of $0.1$?

Tip: When working with summations, double-check each term's calculation to ensure accuracy in the final sum.