Question

Consider a sequence of real numbers $x_1,x_2,x_3,…$ such that $x_{n+1}=x_n+n−1$ for all $n≥1$. If $x_1=−1$ then $x_{100}$ is equal to

• A. 4949
• B. 4849
• C. 4850
• D. 4950

Answer 1

Answer: (C)

4850

Answer 2

xn+1 = xn + n - 1
x 1 = -1
x 2 = x1 + (1 - 1)
x 3 = x2 + (2 - 1)
⇒ x3 = x1 + (1 - 1) + (2 - 1)
x 4 = x3 + (3 - 1)
⇒ x4 = x1 + (1 - 1) + (2 - 1) + (3 - 1)

so, as we see that,
x100 = x1 + (1 - 1) + (2 - 1) + (3 - 1) + … + (99 - 1)
x100 = x1 + (1 + 2 + 3 + 4 + … + 98 + 99) - 99 (1)
x100 = x1 + (1 + 2 + 3 + 4 + … + 98)
x100 = (-1) + ($98×\frac{99}{2}$)
x100 = (-1) + 4851
x100 = 4850