We have: x1−x2+x3−x4+⋯+(−1)n+1xn=n2+2n
This equation tells us the alternating sum of the first n terms equals n2+2n.
Let's find individual terms by substituting small values of n:
For n=1:
x1=12+2(1)=3
For n=2:
x1−x2=22+2(2)=8
Since x1=3: 3−x2=8
Therefore: x2=−5
For n=3:
x1−x2+x3=32+2(3)=15
Substituting: 3−(−5)+x3=15
8+x3=15
Therefore: x3=7
For n=4:
x1−x2+x3−x4=42+2(4)=24
Substituting: 3+5+7−x4=24
15−x4=24
Therefore: x4=−9
Now checking consecutive pairs:
x1+x2=3+(−5)=−2
x3+x4=7+(−9)=−2
Pattern: Each pair of consecutive terms sums to −2.
Since 49 and 50 are consecutive terms:
x49+x50=−2