Question

[Nguyễn Hoàng Sơn]

tính giá trị của biểu thức \(\frac{1}{1+2}\)+\(\frac{1}{1+2+3}\)+\(\frac{1}{1+2+3+4}\)+...+\(\frac{1}{1+2+3+...+99}\)+\(\frac{1}{50}\)

Answer 1

[kagamine rin len]

1/1+2+1/1+2+3+1/1+2+3+4+...+1/1+2+3+...+99 +1/50

=1/(2+1).2:2+1/(3+1).3:2+1/(4+1).4:2+..+1/(99+1).99:2+1/50

=2/2.3+2/3.4+2/4.5+..+2/99.100+1/50

=2(1/2.3+1/3.4+1/4.5+..+1/99.100)+1/50

=2(1/2-1/3+1/3-1/4+1/4-1/5+...+1/99-1/100)+1/50

=2(1/2-1/100)+1/50

=49/50+1/50=1