Question

[Nguyễn Ngọc Quế Anh]

1 Tính tổng sau:

a. 1 + 2 + 3 +4 + ...+ n

b. 2 + 4 + 6 + 8 +...+ 2n

c .1 + 3 + 5 +...+ (2n+1)

d, 1 + 4 + 7 + 10 + ...+ 2008

Answer 1

[Ka anata no kokoro no tame ni]

a) =\(\frac{n\left(n+1\right)}{2}\)

b) =\(n\left(n+1\right)\)

c) =\(\left(n+1\right)^2\)

d) =\(\left(2008+1\right).\left(\frac{2008-1}{3}+1\right):2=673015\)