Consider the sequence t1=1, t2=-1, and that tn=t(n-2)*(n-3)/(n-1), where n>2. Find the value of t1998. (1998 is the term, so is n, t1, and t2)
Did you take a look at the terms?
1, 1, 0, 0, ...
To find the value of t1998 in the given sequence, we need to calculate the terms of the sequence step by step until we reach t1998 using the recursive formula: tn = t(n-2) * (n-3) / (n-1), where n > 2.
Let's calculate the terms of the sequence one by one until we reach t1998:
t1 = 1
t2 = -1
t3 = t(3-2) * (3-3) / (3-1)
= t1 * 0 / 2
= 0 / 2
= 0
t4 = t(4-2) * (4-3) / (4-1)
= t2 * 1 / 3
= -1 * 1 / 3
= -1/3
t5 = t(5-2) * (5-3) / (5-1)
= t3 * 2 / 4
= 0 * 2 / 4
= 0
By observing the pattern, we see that the terms alternate between 0 and -1/3:
t1 = 1
t2 = -1
t3 = 0
t4 = -1/3
t5 = 0
Since the sequence alternates between 0 and -1/3, we can conclude that:
t(3k) = 0 for any positive integer k
t(3k+1) = -1/3 for any positive integer k
To find the value of t1998, we need to determine the remainder when 1998 is divided by 3:
1998 ÷ 3 = 666 with a remainder of 0
Since 1998 is divisible evenly by 3, we conclude that t1998 = t(3k) = 0.
Therefore, t1998 is equal to 0.
To find the value of t1998, we need to calculate the terms of the sequence up to t1998 using the given recursive formula.
We know that t1 = 1 and t2 = -1. We can use these values to generate the rest of the terms step by step.
To find t3:
t3 = t(3-2) * (3-3) / (3-1)
= t1 * 0 / 2
= 0
To find t4:
t4 = t(4-2) * (4-3) / (4-1)
= t2 * 1 / 3
= -1 * 1/3
= -1/3
To find t5:
t5 = t(5-2) * (5-3) / (5-1)
= t3 * 2 / 4
= 0 * 2/4
= 0
Continuing this pattern, we can find that t6 = -1/6, t7 = 0, t8 = -1/10, and so on.
Notice that for even values of n, the term t(n) is always negative, and for odd values of n, the term t(n) is always zero.
Since t1998 is even, we can conclude that t1998 is negative.