3/3.6 + 3/6.9 + 3/9.12 + ... + 3/96.99
\(A=\frac{5}{3.6}+\frac{5}{6.9}+\frac{5}{9.12}+..........+\frac{5}{96.99}\)
3/5 A = 3/3.6 + 3/6.9 +..... + 3/96.99
= 1/3 - 1/6 + 1/6 - 1/9 + .... + 1/96 - 1/99 = 1/3 - 1/99 = 32/99
=> A = 160/297
k mk nha
Tính
a,B=2.4+4.6+6.8+...+98.100
b,B=3+3.6+6.9+...+96.99
a,6B=2.4.6+4.6.(8-2)+...............+98.100.(102-96)
6B=2.4.6+4.6.8-2.4.6+..............+98.100.102-96.98.100
6B=98.100.102
B=98.100.102:6
B=166600
\(F=\frac{2}{3.6}+\frac{2}{6.9}+\frac{2}{9.12}+...+\frac{2}{30.33}+\frac{3}{33.36}\)=?
Tính tổng:
N=2/3.6+2/6.9+2/9.12+....+2/2019.2022
\(N=\dfrac{2}{3.6}+\dfrac{2}{6.9}+...+\dfrac{2}{2019.2022}\)
\(\Rightarrow N=2\left(\dfrac{1}{3.6}+\dfrac{1}{6.9}+...+\dfrac{1}{2019.2022}\right)\)
\(\Rightarrow N=\dfrac{2}{3}\left(\dfrac{3}{3.6}+\dfrac{3}{6.9}+...+\dfrac{3}{2019.2022}\right)\)
\(\Rightarrow N=\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{2019}-\dfrac{1}{2022}\right)\)
\(\Rightarrow N=\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2022}\right)\)
\(\Rightarrow N=\dfrac{2}{3}.\dfrac{673}{2022}\\ \Rightarrow N=\dfrac{673}{3033}\)
1/3.6+1/6.9+1/9.12+1/12.15+1/15.18
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{15}-\dfrac{1}{18}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{18}\right)=\dfrac{1}{3}\cdot\dfrac{5}{18}=\dfrac{5}{54}\)
4/3.6+ 4/6.9+ 4/9.12 + 4/12.15
số hạng của dãy: 1/3.6 ; 1/6.9 ; 1/9.12 ; ... ; 1/156.159
Số số hạng là :
(159 - 6) : 3 + 1 = 52 (số hạng)
Số số hạng là :
(159 - 6) : 3 + 1 = 52 (số hạng)
Chứng minh: 1/3.6 + 1/6.9 + 1/9.12 + ... + 1/219.222 <1
\(S=\frac{1}{3.6}+\frac{1}{6.9}+...+\frac{1}{219.222}\)
\(\Rightarrow3S=\frac{3}{3.6}+\frac{3}{6.9}+...+\frac{3}{219.222}\)
\(=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{219}-\frac{1}{222}\)
\(=\frac{1}{3}-\frac{1}{222}< \frac{1}{3}\)
\(\Rightarrow S< \frac{1}{9}< 1\)
\(\Rightarrow S< 1\left(đpcm\right)\)
Số số hạng của dãy 1/3.6; 1/6.9; 1/9.12;...; 1/156.159
Số số hạng là:(159-6):3+1=52(số hạng)