GreenlightTestPrep wrote:
The sequence of numbers \(t_1\), \(t_2\), \(t_3\), . . . , \(t_n\), . . . is defined by \( t_ n = t_{n-1} - t_{n-2}\) for \(n > 2\).
If \(t_ 1 = 1\) and \(t_ 2 = 1\), What is the sum of the first \(50\) terms of this sequence?
A) \(-2\)
B) \(-1\)
C) \(0\)
D) \(1\)
E) \(2\)
Let’s first list some terms to see if we can spot a pattern…
\(t_ 1 = 1\)
\(t_ 2 = 1\)
\(t_ 3 = t_2 – t_1 = 1 – 1 = 0\)
\(t_ 4 = t_3 – t_2 = 0 – 1 = -1\)
\(t_ 5 = t_4 – t_3 = (-1) – 0 = -1\)
\(t_ 6 = t_5 – t_4 = (-1) – (-1) = 0\)
\(t_ 7 = t_6 – t_5 = 0 - (-1) = 1\)
\(t_ 8 = t_7 – t_6 = 1 - 0 = 1\)
\(t_ 9 = t_8 – t_7 = 1 - 1 = 0\)
\(t_ {10} = t_9 – t_8 = 0 - 1 = -1\)
.
.
.
At this point, we can see that
the pattern repeats itself after 6 terms In other words, the six terms from \(t_1\) to \(t_6\) are exactly the same as the six terms from \(t_7\) to \(t_{12} \) (as well as the six terms from \(t_{13}\) to \(t_{18}\), etc)
The sum of the first six terms \(= 1 + 1 + 0 + (-1) + (-1) + 0 = 0\)
So, the sum of the next six terms after that \(= 0\)
And the sum of the next six terms after that \(= 0\)
Etc.
There are 8 sets of six terms from \(t_1\) to \(t_{48}\), which means the sum of the first 48 terms \(= (8)(0) = 0\)
All we need to do now is add \(t_ {49} \) and \(t_ {50} \)
We can follow the pattern to see that \(t_ {49} = 1\) and \(t_ {50} = 1\), which means the sum of the first 50 terms = 2
Answer: E