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

Question

Let \( a_n \) denote the term independent of \( x \) in the expansion of \[ \left[x + \frac{\sin(1/n)}{x^2}\right]^{3n}, \] then \[ \lim_{n\to\infty} \frac{(a_n)n!}{\,{}^{3n}P_n} \] equals:

Answer 1

The recurrence relation is given as: \( x_{n+1} = x_n + n - 1 \). The initial condition is \( x_1 = -1 \).

Calculating the first few terms:

\( x_2 = x_1 + (1 - 1) = -1 + 0 = -1 \)
\( x_3 = x_2 + (2 - 1) = -1 + 1 = 0 \)
\( x_4 = x_3 + (3 - 1) = 0 + 2 = 2 \)
\( x_5 = x_4 + (4 - 1) = 2 + 3 = 5 \)

The general form can be expressed as a sum:

\[ x_{n} = x_1 + \sum_{k=1}^{n-1}(k - 1) \]

The sum simplifies to:

\[ \sum_{k=1}^{n-1}(k - 1) = \frac{(n - 1)(n - 2)}{2} \]

Thus, the explicit formula for \( x_n \) is:

\[ x_n = x_1 + \frac{(n - 1)(n - 2)}{2} \]

Substituting \( x_1 = -1 \) and \( n = 100 \):

\[ x_{100} = -1 + \frac{(99)(98)}{2} = -1 + 4851 = 4850 \]

Final Answer: \( \boxed{x_{100} = 4850} \)

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

The Correct Option is C

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