Question

a) 1 - 3 + 5 - 7 + 9 - 11 + ... + 2021 - 2023

b) \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{21}\) + \(\dfrac{1}{28}\) + \(\dfrac{1}{36}\) + \(\dfrac{1}{45}\)

Answer 1

a) 1 - 3 + 5 - 7 + 9 - 11 + ... + 2021 - 2023

Số số hạng:

(2023 - 1) : 2 + 1 = 1012 (số)

Số cặp số:

1012 : 2 = 506 (cặp)

Tổng là:

1 - 3 + 5 - 7 + 9 - 11 + ... + 2021 - 2023

= (1 - 3) + (5 - 7) + (9 - 11) + ... + (2021 - 2023)

= -2.506

= -1012

Answer 2

b) Xem lại đề!

Answer 3

Câu b) thiếu \(\dfrac{1}{15}\)

Answer 4

b)

Ta có :

\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}\)

\(=2.\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)

\(=2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\)

\(=2.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=2.\left(\dfrac{1}{2}-\dfrac{1}{10}\right)\)

\(=1-\dfrac{1}{5}=\dfrac{4}{5}\)