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

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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

cdquestions Admin · Accepted Answer

x n+1 = x n + n - 1 x 1 = -1 x 2 = x 1 + (1 - 1) x 3 = x 2 + (2 - 1) ⇒ x 3 = x 1 + (1 - 1) + (2 - 1) x 4 = x 3 + (3 - 1) ⇒ x 4 = x 1 + (1 - 1) + (2 - 1) + (3 - 1) so, as we see that, x 100 = x 1 + (1 - 1) + (2 - 1) + (3 - 1) + … + (99 - 1) x 100 = x 1 + (1 + 2 + 3 + 4 + … + 98 + 99) - 99 (1) x 100 = x 1 + (1 + 2 + 3 + 4 + … + 98) x 100 = (-1) + ( $98×\frac{99}{2}$ ) x 100 = (-1) + 4851 x 100 = 4850

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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

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