Tính tổng M=$ sqrt[2]{1+ frac{1}{1^2}+ frac{1}{2^2}}$+ $ sqrt[2]{1+ frac{1}{2^2}+ frac{1}{3^2}}$+ ...+ $ sqrt[2]{1+ frac{1}{2019^2}+ frac{1}{20

Question

Tính tổng  M=$ sqrt[2]{1+  frac{1}{1^2}+  frac{1}{2^2}}$+ $ sqrt[2]{1+  frac{1}{2^2}+  frac{1}{3^2}}$+ ...+ $ sqrt[2]{1+  frac{1}{2019^2}+  frac{1}{2020^2}}$

cattien · Answer

Đ&aacute;p &aacute;n:      `M = 4080399/2020`     Giải th&iacute;ch c&aacute;c bước giải:     Đặt `A^2 = 1+ 1/a^2 + 1/(a+1)^2= (a^2(a+1)^2+(a+1)^2 + a^2)/(a^2(a+1)^2)`   `= (a^2 (a^2 + 2a + 1 +1)+(a+1)^2)/(a^2(a+1)^2)`   `= (a^4 + 2a^2(a+1)+(a+1)^2)/(a^2(a+1)^2)`   `= (a^2 + a+1)^2/(a^2(a+1)^2)`   `= [(a^2+a+1)/(a(a+1))]^2`   V&igrave; `a>0` n&ecirc;n `A>0` suy ra `A= (a^2 + a+1)/(a(a+1))`   Do đ&oacute; `A = sqrt(1+1/a^2 + 1/(a+1)^2)`   `= (a^2+a+1)/(a(a+1))`   `= 1+1/(a(1+1))`   `= 1+1/a - 1/(a+1)`   Suy ra `M = (1+1/1-1/2) + (1+1/2-1/3)+...+ (1+1/2019-1/2020)`   `= 2019+(1/1-1/2+1/2-1/3+...+1/2019-1/2020)`   `= 2019 + (1-1/2020)`   `= 2020 - 1/2020`   `= (2020^2-1)/2020`   `= 4080399/2020`

Tính tổng M=$\sqrt[2]{1+ \frac{1}{1^2}+ \frac{1}{2^2}}$+ $\sqrt[2]{1+ \frac{1}{2^2}+ \frac{1}{3^2}}$+ …+ $\sqrt[2]{1+ \frac{1}{2019^2}+ \frac{1}{20

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