Đặt A = 1/1.3 + 1/3.5 + 1/5.7 +........+ 1/(2n - 1)(2n + 1)
2.A = 2/1.3 + 2/3.5 + 2/5.7 +........+ 2/(2n - 1)(2n + 1)
2.A = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/(2n - 1) - 1/(2n + 1)
2.A = 1 - 1/(2n + 1) = 2n/(2n + 1)
Vậy A = n/(2n + 1)
Đặt A = 1/1.3 + 1/3.5 + 1/5.7 +........+ 1/(2n - 1)(2n + 1)
2.A = 2/1.3 + 2/3.5 + 2/5.7 +........+ 2/(2n - 1)(2n + 1)
2.A = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/(2n - 1) - 1/(2n + 1)
2.A = 1 - 1/(2n + 1) = 2n/(2n + 1)
Vậy A = n/(2n + 1)
cho \(I=\frac{1.3+2}{4}.\frac{3.5+2}{16}.....\frac{\left(2^{2n}-1\right)\left(2^{2n}+1\right)+2}{2^{2n}}\)với n thuộc N. chứng minh \(I< \frac{4}{3}\)
Bài 10. Tìm số tự nhiên n rhõa mãn :
1/1.3 + 1/3.5 +.......+ 1/( 2n-1)(2n+1) = 2015/4031
Tính : 1.3+3.5+5.7+...+(2n+1).(2n+2)=?
Tính Q=\(\frac{1.3}{3.5}+\frac{2.4}{5.7}+\frac{3.5}{7.9}+.....+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}+......+\frac{1002.1004}{2005.2007}\)
1. 3.5.....(2n-1)/(n+1)(n+2)(n+3) = 1/2n với n thuộc n*
( đề bài là chứng minh rằng )
giúp mình với cảm ơn
Tính: \(Q=\frac{1.3}{3.5}+\frac{2.4}{5.7}+\frac{3.5}{7.9}+...+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}+...+\frac{1002.1004}{2005.2007}\)
Tính: \(Q=\frac{1.3}{3.5}+\frac{2.4}{5.7}+\frac{3.5}{7.9}+...+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}+...+\frac{1002.1004}{2005.2007}\)
chứng minh rằng
1, 1/n(n+1)=1/n-1/n+1
2, 2/n(n+1)(n+2)=1/n(n+1)-1/(n+1)(n+2)
3, 3/n(n+1)(n+2)(n+3)=1/n(n+1)(n+2)-1/(n+1)(n+2)(n+3)
4, 4/(2n-1)(2n+1)(2n+3)=1/(2n+1)(2n-1)-1/(2n+1)(2n+3)
5, m/n(n+m)=1/n-1/n+m
6, 2m/n(n+m)(n+2n)=1/n(n+m)-1/(n+m)(n+2n)
1. tính:
a) 1.3+2.4+3.5+4.6+...+n.(n+2)
b) 1.5+2.6+3.7+...+n.(n+4)
c) 12 + 32+52+...+(2n+1)2