The sequence 1, 4, 13, 40, 121 can be described by a recursive formula. The recursive relationship can be expressed as ( a_n = 3a_{n-1} + 1 ) for ( n \geq 2 ), with the initial condition ( a_1 = 1 ). This means each term is generated by multiplying the previous term by 3 and then adding 1.
121
22
364
To get the next term, triple the present term and add 1 . The next term after 40 is 121 .
1,093 3,2804 - 1 = 313 - 4 = 940 - 13 = 27121 - 40 = 81364 - 121 = 243each difference is 3 times the previous243 x 3 = 729 then 364 + 729 = 1,093729 x 3 = 2,187 then 1,093 + 2,187 = 3,280
121
22
364
To get the next term, triple the present term and add 1 . The next term after 40 is 121 .
The factors of 121 are 1, 11, 121 The factors of 280 are 1, 2, 4, 5, 7, 8, 10, 14, 20, 28, 35, 40, 56, 70, 140, 280
121/40
1,093 3,2804 - 1 = 313 - 4 = 940 - 13 = 27121 - 40 = 81364 - 121 = 243each difference is 3 times the previous243 x 3 = 729 then 364 + 729 = 1,093729 x 3 = 2,187 then 1,093 + 2,187 = 3,280
The number is 55+66 = 121
121 divided by 3 equals 40 with a remainder of 1.
42 + 79 = 121 (40+70) + (2+9) = = 110 + 11 = 121
No
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