Question

Tính

B=1-\(\frac{1}{2}\)(1+2)-\(\frac{1}{3}\)(1+2+3)-\(\frac{1}{4}\)(1+2+3+4)-...-\(\frac{1}{20}\)(1+2+3+...+20)

Answer 1

\(B=1-\frac{1}{2}\left(1+2\right)-\frac{1}{3}.\left(1+2+3\right)-\frac{1}{4}.\left(1+2+3+4\right)-...-\frac{1}{20}.\left(1+2+3+...+20\right)\)

\(B=1-\frac{1}{2}.\left(1+2\right).2:2-\frac{1}{4}.\left(1+4\right).4:2-...-\frac{1}{20}.\left(1+20\right).20:2\)

\(B=1-3:2-5:2-...-21:2\)

\(B=1-3.\frac{1}{2}-5.\frac{1}{2}-...-21.\frac{1}{2}\)

\(B=1-\frac{1}{2}.\left(3+5+...+21\right)\)

Đặt C = 3 + 5 + ... + 21

Số số hạng của tổng C là: (21 - 3) : 2 + 1 = 10 (số)

=> C = (3 + 21) x 10 : 2 = 24 x 5 = 120

=> \(A=1-\frac{1}{2}.120\)

\(A=1-60=-59\)