Cho cấp số cộng (un) với u1=11; u2=13. Tính tổng S=1/u1u2+1/u2u3+…..+1/u99u100 13/10/2021 Bởi Hadley Cho cấp số cộng (un) với u1=11; u2=13. Tính tổng S=1/u1u2+1/u2u3+…..+1/u99u100
Đáp án: $S = \dfrac{9}{{209}}$ Giải thích các bước giải: $\begin{array}{l}{u_2} – {u_1} = 2 \Rightarrow d = 2\\S = \dfrac{1}{{{u_1}.{u_2}}} + \dfrac{1}{{{u_2}{u_3}}} + .. + \dfrac{1}{{{u_{99}}.{u_{100}}}}\\ \Rightarrow S = \dfrac{1}{2}.\left( {\dfrac{2}{{{u_1}.{u_2}}} + \dfrac{2}{{{u_2}.{u_3}}} + … + \dfrac{2}{{{u_{99}}.{u_{100}}}}} \right)\\ \Rightarrow S = \dfrac{1}{2}.\left( {\dfrac{1}{{{u_1}}} – \dfrac{1}{{{u_2}}} + \dfrac{1}{{{u_2}}} – \dfrac{1}{{{u_3}}} + … + \dfrac{1}{{{u_{99}}}} – \dfrac{1}{{{u_{100}}}}} \right)\\ \Rightarrow S = \dfrac{1}{2}.\left( {\dfrac{1}{{{u_1}}} – \dfrac{1}{{{u_{100}}}}} \right)\\Do:{u_{100}} = {u_1} + 99.d = 11 + 99.2 = 209\\ \Rightarrow S = \dfrac{1}{2}.\left( {\dfrac{1}{{11}} – \dfrac{1}{{209}}} \right) = \dfrac{9}{{209}}\end{array}$ Bình luận
Đáp án: $S = \dfrac{9}{{209}}$
Giải thích các bước giải:
$\begin{array}{l}
{u_2} – {u_1} = 2 \Rightarrow d = 2\\
S = \dfrac{1}{{{u_1}.{u_2}}} + \dfrac{1}{{{u_2}{u_3}}} + .. + \dfrac{1}{{{u_{99}}.{u_{100}}}}\\
\Rightarrow S = \dfrac{1}{2}.\left( {\dfrac{2}{{{u_1}.{u_2}}} + \dfrac{2}{{{u_2}.{u_3}}} + … + \dfrac{2}{{{u_{99}}.{u_{100}}}}} \right)\\
\Rightarrow S = \dfrac{1}{2}.\left( {\dfrac{1}{{{u_1}}} – \dfrac{1}{{{u_2}}} + \dfrac{1}{{{u_2}}} – \dfrac{1}{{{u_3}}} + … + \dfrac{1}{{{u_{99}}}} – \dfrac{1}{{{u_{100}}}}} \right)\\
\Rightarrow S = \dfrac{1}{2}.\left( {\dfrac{1}{{{u_1}}} – \dfrac{1}{{{u_{100}}}}} \right)\\
Do:{u_{100}} = {u_1} + 99.d = 11 + 99.2 = 209\\
\Rightarrow S = \dfrac{1}{2}.\left( {\dfrac{1}{{11}} – \dfrac{1}{{209}}} \right) = \dfrac{9}{{209}}
\end{array}$