Hình như bạn viết thiếu \(\frac{1}{36}\), nếu đúng là vậy thì mình giải như sau.
Đặt A =\(\frac{1}{6}+\frac{1}{18}+\frac{1}{36}+\frac{1}{60}+\frac{1}{90}+\frac{1}{126}\)
3A = \(\frac{3}{3.2}+\frac{3}{3.6}+\frac{3}{3.12}+\frac{3}{3.20}+\frac{3}{3.30}+\frac{3}{3.42}\)
3A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
3A = \(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{7-6}{6.7}\)
3A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\)
3A = \(1-\frac{1}{7}=\frac{6}{7}\)
A = \(\frac{6}{7}:3=\frac{2}{7}\)
Gọi tổng là A:
ta có :
3A= 1/2+1/6+1/20+1/30+1/42
3A= 1/1.2+1/2.3+1/4.5+1/6.7
3A= 1-1/2+1/2-1/3+1/4-1/5+1/6-1/7
3A=1-1/3+1/4-1/5+1/6-1/7
3A= (1-1/7)+(1/3-1/4)+(1/5-1/7)
3A=6/7+-1/12+-2/35
3A=6/7-2/35+-1/12
3A=4/5-1/12
3A=43/60
A=43/180
