Question

Bài 1 : Tính  : C = 1 + 1/2 ( 1 + 2 ) + 1/3 ( 1+2+3) +... + 1/16 (1+2+3+...+16 )

Answer 1

\(C=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)

\(C=1+\frac{1}{2}.\left(1+2\right).2:2+\frac{1}{3}.\left(1+3\right).3:2+...+\frac{1}{16}.\left(1+16\right).16:2\)

\(C=2:2+3:2+4:2+...+17:2\)

\(C=2.\frac{1}{2}+3.\frac{1}{2}+4.\frac{1}{2}+...+17.\frac{1}{2}\)

\(C=\frac{1}{2}.\left(2+3+4+...+17\right)\)

\(C=\frac{1}{2}.\left(2+17\right).16:2\)

\(C=\frac{1}{2}.19.16.\frac{1}{2}\)

\(C=19.4=76\)