Consider a sequence of real numbers, $$x_{1},x_{2},x_{3},...$$ such that $$x_{n+1}=x_{n}+n-1$$ for all $$n\geq1$$. If $$x_{1}=-1$$ then $$x_{100}$$ is equal to
GivenĀ $$x_{n+1}\ =\ x_n\ +\ n\ -1$$ and x1 = -1.
ConsideringĀ
x1Ā Ā = -1.Ā Ā Ā (1)
x2Ā Ā = x1+1-1 = x1 + 0Ā Ā (2)
x3Ā Ā = x2 + 2 - 1Ā =x2 + 1Ā Ā Ā (3)
x4Ā Ā = x3 + 3 - 1 = x3 + 2Ā Ā Ā Ā (4)
x100 = x99 + 98Ā Ā Ā (100)
Adding the LHS and RHS for the hundred equations we have:
(x1+x2+......................x100) = (-1+0+.........98) + (x1+x2+...............x99)
Subtracting this we have :
(x1+...........x100) - (x1+............. x 99) = $$\frac{\left(98\cdot99\right)}{2}$$ - 1.
x100 = 4851 - 1 = 4850
Alternatively
$$x_1=-1$$
$$x_2=x_1+1-1=x_1=-1$$
$$x_3=x_2+2-1=x_2+1=-1+1=0$$
$$x_4=x_3+3-1=x_3+2=0+2=2$$
$$x_5=x_4+4-1=x_4+3=2+3=5$$
......
If we observe the series, it is a series that hasĀ a difference between the consecutive terms in an AP.
Such series are represented asĀ $$t\left(n\right)=a+bn+cn^2$$
We need to find t(100).
t(1) = -1
a + b + c = -1
t(2) = -1
a + 2b + 4c = -1
t(3) = 0
a + 3b + 9c = 0
Solving we get,
b + 3c = 0
b + 5c = 1
c = 0.5
b = -1.5
a = 0
Now,Ā
$$t\left(100\right)=\left(-1.5\right)100+\left(0.5\right)100^2=-150+5000=4850$$
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