`3/(1^2. 2^2) + 5/(2^2 . 3^2) + 7/(3^2. 4^2)+…+ 19/(9^2. 10^2)`

1 bình luận về “`3/(1^2. 2^2) + 5/(2^2 . 3^2) + 7/(3^2. 4^2)+…+ 19/(9^2. 10^2)`”

1. Giải đáp: $\dfrac{99}{100}$

    

   Lời giải và giải thích chi tiết:

   $\dfrac{3}{{{1}^{2}}{{.2}^{2}}}+\dfrac{5}{{{2}^{2}}{{.3}^{2}}}+\dfrac{7}{{{3}^{2}}{{.4}^{2}}}+…+\dfrac{19}{{{9}^{2}}{{.10}^{2}}}$

   $=\dfrac{{{2}^{2}}-{{1}^{2}}}{{{1}^{2}}{{.2}^{2}}}+\dfrac{{{3}^{2}}-{{2}^{2}}}{{{2}^{2}}{{.3}^{2}}}+\dfrac{{{4}^{2}}-{{3}^{2}}}{{{3}^{2}}{{.4}^{2}}}+…+\dfrac{{{10}^{2}}-{{9}^{2}}}{{{9}^{2}}{{.10}^{2}}}$

   $=\dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{2}^{2}}}+\dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{3}^{2}}}+\dfrac{1}{{{3}^{2}}}-\dfrac{1}{{{4}^{2}}}+…+\dfrac{1}{{{9}^{2}}}-\dfrac{1}{{{10}^{2}}}$

   $=\dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{10}^{2}}}$

   $=\dfrac{{{10}^{2}}-{{1}^{2}}}{{{1}^{2}}{{.10}^{2}}}$

   $=\dfrac{99}{100}$

   Trả lời