i0 = 4; in = in-1 - 3
Example 1:+5 - +2Step 1:The opposite of +2 is -2.Step 2:Subtraction becomes addition.Solution:+5 - +2 = +5 + -2 = +3
n*3; n-1 {Where "n" is the previous answer starting at 1. 1, 1*3=3, 3-1=2, 2*3=6, 6-1=5, 5*3=15, 15-1=14, 14*3=42, 42-1=41.
3 / 5 * 5 / 3 = 1 This answer was calculated according to the order of operations. Rule 1: First perform any calculations inside parentheses. Rule 2: Next, from left to right, do all multiplications and divisions. Rule 3: Last, from left to right, do all additions and subtractions.
1 2 3 4 5 2 5 8 11 14 ... If this is the sequence, the position-to-term rule is 3n-1. However, it could be another sequence depending on the rest of the terms.
the rule is plus 2 1/3
i0 = 4; in = in-1 - 3
Example 1:+5 - +2Step 1:The opposite of +2 is -2.Step 2:Subtraction becomes addition.Solution:+5 - +2 = +5 + -2 = +3
If you mean 2/1 3/2 4/3 5/4 then the next 3 terms are 6/5 7/6 8/7 and the nth term is (n+1)/n
n*3; n-1 {Where "n" is the previous answer starting at 1. 1, 1*3=3, 3-1=2, 2*3=6, 6-1=5, 5*3=15, 15-1=14, 14*3=42, 42-1=41.
3 / 5 * 5 / 3 = 1 This answer was calculated according to the order of operations. Rule 1: First perform any calculations inside parentheses. Rule 2: Next, from left to right, do all multiplications and divisions. Rule 3: Last, from left to right, do all additions and subtractions.
1 2 3 4 5 2 5 8 11 14 ... If this is the sequence, the position-to-term rule is 3n-1. However, it could be another sequence depending on the rest of the terms.
There are 64 subsets, and they are:{}, {A}, {1}, {2}, {3}, {4}, {5}, {A,1}, {A,2}, {A,3}, {A,4}, {A,5}, {1,2}, {1,3}, {1,4}, {1,5}, {2,3}, {2,4}, {2,5}, {3,4}, {3, 5}, {4,5}, {A, 1, 2}, {A, 1, 3}, {A, 1, 4}, {A, 1, 5}, {A, 2, 3}, {A, 2, 4}, {A, 2, 5}, {A, 3, 4}, {A, 3, 5}, {A, 4, 5}, {1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, 5}, {2, 3, 4}, {2, 3, 5}, {2, 4, 5}, {3, 4, 5}, {A, 1, 2, 3}, {A, 1, 2, 4}, {A, 1, 2, 5}, {A, 1, 3, 4}, {A, 1, 3, 5}, {A, 1, 4, 5}, {A, 2, 3, 4}, {A, 2, 3, 5}, {A, 2, 4, 5}, {A, 3, 4, 5}, {1, 2, 3, 4}, {1, 2, 3, 5}, {1, 2, 4, 5}, {1, 3, 4, 5}, {2, 3, 4, 5}, {A, 1, 2, 3, 4}, {A, 1, 2, 3, 5}, {A, 1, 2, 4, 5}, {A, 1, 3, 4, 5}, {A, 2, 3, 4, 5}, {1, 2, 3, 4, 5} {A, 1, 2, 3,,4, 5} .
+1, +2, +3, +4, +5, +6, +7, +8
The rule is t(n) = 5 + 2*n, where n = 1, 2, 3, ...
Add the previous 2 numbers to get the next number.
for 3: count all number if divisable by 3 (123= 1+2+3= 6, 3+3=6) for 5:end number is 5 or 0