Question

[Hạnh Nguyên]

Tính: 1/(1+2+3) + 1/(1+2+3+4) + 1/(1+2+3+4+5) + ... + 1/(1+2+3+...+2023)

Answer 1

[Akai Haruma]

Lời giải:
Gọi tổng trên là $A$
$A=\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+....+\frac{1}{\frac{2023.2024}{2}}$

$=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2023.2024}$

$=2(\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{2024-2023}{2023.2024})$

$=2(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{2023}-\frac{1}{2024})$

$=2(\frac{1}{3}-\frac{1}{2024})=\frac{2021}{3036}$

Answer 2

[Dũng Đậu]

$A=\genfrac{}{}{0ex}{}{1}{\genfrac{}{}{0ex}{}{3.4}{2}}+\genfrac{}{}{0ex}{}{1}{\genfrac{}{}{0ex}{}{4.5}{2}}+....+\genfrac{}{}{0ex}{}{1}{\genfrac{}{}{0ex}{}{2023.2024}{2}}$

$=\genfrac{}{}{0ex}{}{2}{3.4}+\genfrac{}{}{0ex}{}{2}{4.5}+...+\genfrac{}{}{0ex}{}{2}{2023.2024}$

$=2(\genfrac{}{}{0ex}{}{4-3}{3.4}+\genfrac{}{}{0ex}{}{5-4}{4.5}+...+\genfrac{}{}{0ex}{}{2024-2023}{2023.2024})$

$=2(\genfrac{}{}{0ex}{}{1}{3}-\genfrac{}{}{0ex}{}{1}{4}+\genfrac{}{}{0ex}{}{1}{4}-\genfrac{}{}{0ex}{}{1}{5}+....+\genfrac{}{}{0ex}{}{1}{2023}-\genfrac{}{}{0ex}{}{1}{2024})$

$=2(\genfrac{}{}{0ex}{}{1}{3}-\genfrac{}{}{0ex}{}{1}{2024})=\genfrac{}{}{0ex}{}{2021}{3036}$