Answer :
D \(\frac{100}{101}\)

Explanation:

Given
a_k = \(\frac{1}{k}\)-\(\frac{1}{k+1}\);

So the sum of a_1 to a_100 is explained below-
a_1 + a_2 + a_3 + ...... + a_100
= \(\frac{1}{1}\) - \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\)+ ..... + \(\frac{1}{100}\) - \(\frac{1}{101}\)

= 1- \(\frac{1}{101}\)

= \(\frac{100}{101}\)
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