Question

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

Answer 1

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

A=23.4​1​+24.5​1​+....+22023.2024​1​

=23.4+24.5+...+22023.2024=3.42​+4.52​+...+2023.20242​

=2(4−33.4+5−44.5+...+2024−20232023.2024)=2(3.44−3​+4.55−4​+...+2023.20242024−2023​)

=2(13−14+14−15+....+12023−12024)=2(31​−41​+41​−51​+....+20231​−20241​)

=2(13−12024)=20213036=2(31​−20241​)=30362021​