Question

M = 1+ 1/2 . ( 1+2) + 1/3.(1+2+3)+....+ 1/99.(1+2+3+...+99)

Answer 1

\(M=1+\frac{1}{2.\left(1+2\right)}+\frac{1}{3.\left(1+2+3\right)}+...+\frac{1}{99.\left(1+2+3+...+99\right)}\)

\(M=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{4950}\)

\(M=1-\frac{1}{4950}\)

\(M=\frac{4949}{4950}\)

Answer 2

mk sai r

Answer 3

\(M=\frac{3}{2}-\frac{1}{4950}=\frac{7424}{4950}\)