Roots!

Consider $i$ -th term,

$\sqrt{1 + \frac{1}{i^2} + \frac{1}{(i+1)^2}} = \sqrt{\frac{i^4 + 2i^3 + 3i^2 + 2i + 1}{i^2(i+1)^2}} = 1 + \frac{1}{i(i+1)} = 1 + \frac{1}{i} - \frac{1}{i+1}$

By telescoping, we get $98 + \frac{1}{2} - \frac{1}{100} = 98 + \frac{49}{100} = 98.49$ .

Or, alternatively, calculate it through python.

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>>> import math
>>> print sum([math.sqrt(1 + 1.0/i**2 + 1.0/(i+1)**2) for i in xrange(2, 100)])
98.94