Let |zr−r| ≤ r, ∀ r =1, 2, 3, ..., n. Then |∑r=1nzr| is less than

|∑r=1nzr| ≤ ∑r=1n|zr| ≤ ∑r=1nzr-r+r≤ ∑r=1n|zr−r|+∑r=1nr ≤ 2∑r=1nr =2nn+12=nn+1 since z1+z2≤z1 +z2

Let |zr−r| ≤ r, ∀ r =1, 2, 3, ..., n. Then |∑r=1nzr| is less than

|∑r=1nzr| ≤ ∑r=1n|zr| ≤ ∑r=1nzr-r+r≤ ∑r=1n|zr−r|+∑r=1nr ≤ 2∑r=1nr =2nn+12=nn+1 since z1+z2≤z1 +z2

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Let $| z_{r} - r | \leq r, \forall r = 1, 2, 3, ..., n .$ Then $| \sum_{r = 1}^{n} z_{r} |$ is less than

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$n$

$2 n$

$n (n + 1)$

$\frac{n (n + 1)}{2}$

answer is C.

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Detailed Solution

$| \sum_{r = 1}^{n} z_{r} | \leq \sum_{r = 1}^{n} | z_{r} | \leq \sum_{r = 1}^{n} |(z_{r} - r) + r| \leq \sum_{r = 1}^{n} | z_{r} - r | + \sum_{r = 1}^{n} r \leq 2 \sum_{r = 1}^{n} r = 2 \frac{n (n + 1)}{2} = n (n + 1) (\sin c e |z_{1} + z_{2}| \leq |z_{1}| + |z_{2}|)$