GreenlightTestPrep wrote:
The sequence of numbers t1, t2, t3, . . . , tn, . . . is defined by tn=tn−1−tn−2 for n>2.
If t1=1 and t2=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…
t1=1t2=1t3=t2–t1=1–1=0t4=t3–t2=0–1=−1t5=t4–t3=(−1)–0=−1t6=t5–t4=(−1)–(−1)=0t7=t6–t5=0−(−1)=1t8=t7–t6=1−0=1t9=t8–t7=1−1=0t10=t9–t8=0−1=−1.
.
.
At this point, we can see that
the pattern repeats itself after 6 terms In other words, the six terms from
t1 to
t6 are exactly the same as the six terms from
t7 to
t12 (as well as the six terms from
t13 to
t18, etc)
The sum of the first six terms
=1+1+0+(−1)+(−1)+0=0So, the sum of the next six terms after that
=0And the sum of the next six terms after that
=0Etc.
There are 8 sets of six terms from
t1 to
t48, which means the sum of the first 48 terms
=(8)(0)=0All we need to do now is add
t49 and
t50 We can follow the pattern to see that
t49=1 and
t50=1, which means the sum of the first 50 terms = 2
Answer: E