Question

小学题目 帮帮忙 1/（1+2+3）+1/（2+3+4）+1/（3+4+5）+…+1/（9+

小学题目 帮帮忙 1/（1+2+3）+1/（2+3+4）+1/（3+4+5）+…+1/（9+10+11）=

Answer 1

分母为等差数列，公差为3
原式=1/3(1/2+1/3+1/4+1/5+1/6+1/7+1/8+1/9+1/10)
=1/3【(1/2+1/3+1/6）+1/4+1/8+1/5+1/10+1/7+1/9】
=1/3（1+3/8+3/10+16/63）=4261/5760

Answer 2

原式=1/6+1/(1×3+6)+1/(2×3+6)+1/(3×3+6)+…1/(8×3+6)
=(1/3)[1/2+1/(1+2)+1/(2+2)+1/(3+2)+…+1/(8+2)]
=(1/3)(1/2+1/3+1/4+…+1/10)
=(1/3)[(1/2+1/3+1/6)+(1/4+1/8)+(1/5+1/10)+(1/7+1/9)]
=(1/3)(1+3/8+3/10+16/63)
=1/3+1/8+1/10+16/189
=1/3+9/40+16/189
=1/3+2341/7560
=4861/7560

Answer 3

=1/(2*3)+1/(3*3)+1/(4*3)+...+1/(10*3)
=1/2*1/3+1/3*1/3+1/4*1/3+...+1/10*1/3
=1/3*(1/2+1/3+1/4+...+1/10)
=4261/5760