Let's analyze the problem step by step: --- ## Given Data: - \( \{a_n\}_{n=} \) is a sequence. - \( s_n = \sum_{k=}^{n} a_k \). - \( s_n = 4 \left(\frac{5}{6}\right)^n \). --- ## Objective: Determine \( a_4 + a_5 + a_6 \). --- ## Step 1: Express \( a_n \) in terms of \( s_n \) Recall: \[ a_n = s_n - s_{n-1} \] for \( n \geq 1 \). --- ## Step 2: Find \( a_n \) explicitly Given: \[ s_n = 4 \left(\frac{5}{6}\right)^n \] and \[ s_{n-1} = 4 \left(\frac{5}{6}\right)^{n-1} \] Therefore: \[ a_n = s_n - s_{n-1} = 4 \left(\frac{5}{6}\right)^n - 4 \left(\frac{5}{6}\right)^{n-1} \] Factor out common terms: \[ a_n = 4 \left(\frac{5}{6}\right)^{n-1} \left(\frac{5}{6} - 1\right) \] \[ a_n = 4 \left(\frac{5}{6}\right)^{n-1} \left(-\frac{1}{6}\right) \] \[ a_n = -\frac{4}{6} \left(\frac{5}{6}\right)^{n-1} \] Simplify: \[ a_n = -\frac{2}{3} \left(\frac{5}{6}\right)^{n-1} \] --- ## Step 3: Calculate \( a_4, a_5, a_6 \) - \( a_4 = -\frac{2}{3} \left(\frac{5}{6}\right)^{3} \) - \( a_5 = -\{2}{3} \left(\frac{5}{6}\right)^{4} \) - \( a6 = -\frac{2}{3} \left(\frac{5}{6}\right)^{5} \) --- ## Step 4: Sum \( a_4 + a_5 + a_6 \) \[ a_4 + a_5 + a_6 = -\frac{2}{3} \left[ \left(\frac{5}{6}\right)^3 + \left(\frac{5}{6}\right)^4 + \left(\frac{5}{6}\right)^5 \right] \] Factor out the common term \( \left(\frac{5}{6}\right)^3 \): \[ a_4 + a_5 + a_6 = -\frac{2}{3} \left(\frac{5}{6}\right)^3 \left[1 + \frac{5}{6} + \left(\frac{5}{6}\right)^2 \right] \] Calculate each term: \[ \left(\frac{5}{6}\right)^3 = \frac{125}{216} \] \[ \left(\frac{5}{6}\right)^2 = \frac{25}{36} \] Sum inside brackets: \[ 1 + \frac{5}{6} + \frac{25}{36} \] Express all over denominator 36: \[ \frac{36}{36} + \frac{30}{36} + \frac{25}{36} = \frac{36 + 30 + 25}{36} = \frac{91}{36} \] Now, combine everything: \[ a_4 + a_5 + a_6 = -\frac{2}{3} \times \frac{125}{216} \times \frac{91}{36} \] Simplify step by step: - Multiply numerators: \[ 2 \times 125 \times 91 = 2 \times 11375 = 22750 \] - Multiply denominators: \[ 3 \times 216 \times 36 \] Calculate denominator: \[ 3 \times 216 = 648 \] \[ 648 \times 36 = 23328 \] Thus: \[ a_4 + a_5 + a_6 = - \frac{22750}{23328} \] --- ## **Final Answer:** \[ \boxed{ a_4 + a_5 + a_6 = - \frac{22750}{23328} } \]

# Problem Overview In this problem, we are tasked with calculating the sum \( a_4 + a_5 + a_6 \) from a sequence defined by the partial sums \( s_n = 4 \left(\frac{5}{6}\right)^n \). --- ## Given Information - Sequence: \( \{a_n\} \) - Partial Sums: \( s_n = \sum_{k=0}^{n} a_k \) - Form of Partial Sums: \( s_n = 4 \left(\frac{5}{6}\right)^n \) --- ## Step-by-Step Solution ### Step 1: Express \( a_n \) The individual terms of the sequence can be calculated using the formula: \[ a_n = s_n - s_{n-1} \] ### Step 2: Calculate \( s_{n-1} \) Given: \[ s_n = 4 \left(\frac{5}{6}\right)^n \] We can express \( s_{n-1} \) as: \[ s_{n-1} = 4 \left(\frac{5}{6}\right)^{n-1} \] ### Step 3: Substitute and Simplify \( a_n \) Now, we find \( a_n \): \[ a_n = s_n - s_{n-1} = 4 \left(\frac{5}{6}\right)^n - 4 \left(\frac{5}{6}\right)^{n-1} \] Factor out common terms: \[ a_n = 4 \left(\frac{5}{6}\right)^{n-1} \left(\left(\frac{5}{6}\right) - 1\right) \] This simplifies to: \[ a_n = 4 \left(\frac{5}{6}\right)^{n-1} \left(-\frac{1}{6}\right) = -\frac{4}{6} \left(\frac{5}{6}\right)^{n-1} \] Further simplification gives: \[ a_n = -\frac{2}{3} \left(\frac{5}{6}\right)^{n-1} \] ### Step 4: Calculate \( a_4, a_5, a_6 \) Using the formula derived: - **For \( a_4 \)**: \[ a_4 = -\frac{2}{3} \left(\frac{5}{6}\right)^{3} \] - **For \( a_5 \)**: \[ a_5 = -\frac{2}{3} \left(\frac{5}{6}\right)^{4} \] - **For \( a_6 \)**: \[ a_6 = -\frac{2}{3} \left(\frac{5}{6}\right)^{5} \] ### Step 5: Sum \( a_4 + a_5 + a_6 \) We proceed to calculate the total: \[ a_4 + a_5 + a_6 = -\frac{2}{3} \left[ \left(\frac{5}{6}\right)^{3} + \left(\frac{5}{6}\right)^{4} + \left(\frac{5}{6}\right)^{5} \right] \] Factor out the common term \( \left(\frac{5}{6}\right)^{3} \): \[ = -\frac{2}{3} \left(\frac{5}{6}\right)^{3} \left[ 1 + \frac{5}{6} + \left(\frac{5}{6}\right)^{2} \right] \] ### Step 6: Calculate the sum inside brackets Calculate: \[ 1 + \frac{5}{6} + \left(\frac{5}{6}\right)^{2} \] Convert to a common denominator of 36: \[ 1 = \frac{36}{36}, \quad \frac{5}{6} = \frac{30}{36}, \quad \left(\frac{5}{6}\right)^{2} = \frac{25}{36} \] Now sum: \[ \frac{36 + 30 + 25}{36} = \frac{91}{36} \] ### Step 7: Final Calculation Substituting back into our sum: \[ a_4 + a_5 + a_6 = -\frac{2}{3} \left(\frac{5}{6}\right)^{3} \cdot \frac{91}{36} \] Calculating \( \left(\frac{5}{6}\right)^{3} = \frac{125}{216} \): Now, substitute: \[ = -\frac{2}{3} \cdot \frac{125}{216} \cdot \frac{91}{36} \] ### Step 8: Combine Numerators and Denominators **Numerators**: \[ 2 \times 125 \times 91 = 22750 \] **Denominators**: Calculating: \[ 3 \times 216 \times 36 = 23328 \] ### Final Answer Thus, we arrive at: \[ \boxed{ a_4 + a_5 + a_6 = -\frac{22750}{23328} } \]