Cho một cấp số cộng (un) có u1

D

- Hướng dẫn giải

Ta có \({S_{100}} = 24850 \Leftrightarrow \frac{n}{2}\left( {{u_1} + {u_n}} \right) = 24850 \Leftrightarrow {u_{100}} = 496\).

Vậy \({u_{100}} = {u_1} + 99d \Leftrightarrow d = \frac{{{u_{100}} - {u_1}}}{{99}} \Leftrightarrow d = 5\).

\(\begin{array}{l}S = \frac{1}{{u_1^{}{u_2}}} + \frac{1}{{{u_2}{u_3}}} + ... + \frac{1}{{{u_{49}}{u_{50}}}}\\ = \frac{1}{{1.6}} + \frac{1}{{6.11}} + \frac{1}{{11.16}} + ... + \frac{1}{{241.246}}\\ \Rightarrow 5S = \frac{5}{{1.6}} + \frac{5}{{6.11}} + \frac{5}{{11.16}} + ... + \frac{5}{{241.246}}\\ = \frac{1}{1} - \frac{1}{6} + \frac{1}{6} - \frac{1}{{11}} + ... + \frac{1}{{241}} - \frac{1}{{246}}\\ = \frac{1}{1} - \frac{1}{{246}} = \frac{{245}}{{246}} \Rightarrow S = \frac{{49}}{{246}}\end{array}\)

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