Today we'll explore a special type of sequence called a jump sequence. A sequence is defined as a jump sequence if for any three consecutive terms, they satisfy the relation: a subscript n minus 1 plus a subscript n plus 1 equals 2 times a subscript n plus negative 1 to the power n. This condition creates an alternating pattern that makes the sequence 'jump' in a specific mathematical way.
Let's examine part 1 of the problem. We need to determine if the arithmetic sequence a subscript n equals 2n minus 1 is a jump sequence. First, we calculate the consecutive terms: a subscript n minus 1 equals 2n minus 3, a subscript n equals 2n minus 1, and a subscript n plus 1 equals 2n plus 1. Now we check the jump condition. The left side gives us a subscript n minus 1 plus a subscript n plus 1 equals 4n minus 2. The right side gives us 2 times a subscript n plus negative 1 to the power n, which equals 4n minus 2 plus negative 1 to the power n. Since these are not equal, the arithmetic sequence is not a jump sequence.
Now let's examine the geometric sequence a subscript n equals 2 to the power n. We calculate the consecutive terms: a subscript n minus 1 equals 2 to the power n minus 1, a subscript n equals 2 to the power n, and a subscript n plus 1 equals 2 to the power n plus 1. For the jump condition, the left side gives us 2 to the power n minus 1 plus 2 to the power n plus 1. This simplifies to 2 to the power n divided by 2 plus 2 times 2 to the power n, which equals 5 times 2 to the power n divided by 2. The right side gives us 2 times 2 to the power n plus negative 1 to the power n, which equals 2 to the power n plus 1 plus negative 1 to the power n. Since these expressions are not equal, the geometric sequence is also not a jump sequence.
Now we move to part 2 of the problem. We have a sequence b subscript n that satisfies the recurrence relation b subscript n plus 1 minus b subscript n equals negative 1 to the power n. We need to prove that this sequence being a jump sequence is equivalent to b subscript 1 equals 1. First, let's find the general term. Using the telescoping method, we can write b subscript 2 minus b subscript 1 equals negative 1, b subscript 3 minus b subscript 2 equals 1, and so on. Summing these differences, we get b subscript n equals b subscript 1 plus the sum from k equals 1 to n minus 1 of negative 1 to the power k. This gives us an alternating pattern where b subscript n equals b subscript 1 for odd n, and b subscript n equals b subscript 1 minus 1 for even n.