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

The Correct Option is C

x_n+1 = x_n + n - 1
x₁ = -1
x₂ = x₁ + (1 - 1)
x₃ = x₂ + (2 - 1)
⇒ x₃ = x₁ + (1 - 1) + (2 - 1)
x₄ = 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