Solve dp[i][c] = dp[i - 1][c] - Socratic AI (ScanMath)

求值

d p i c = d p (i - 1) c

使用可交换属性对术语进行重新排序

i d p c = d p (i - 1) c

简化

更多步骤

i d p c = (- 1 + i) d p c

将表达式移到左侧

i d p c - (- 1 + i) d p c = 0

把各项相减

更多步骤

d p c = 0

将方程分成 3 种可能的情况

d = 0 p = 0 c = 0

求集合的并集

c = 0 d = 0 p = 0

解题方案

(c_{1}, d_{1}, p_{1}) = (c, 0, p), (c, p) \in R^{2} (c_{2}, d_{2}, p_{2}) = (c, d, 0), (c, d) \in R^{2} (c_{3}, d_{3}, p_{3}) = (0, d, p), (d, p) \in R^{2}

Dp[i][c] = dp[i 1][c]