\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)

\(=\frac{1}{2}-\frac{1}{20}\)

\(=\frac{9}{20}\)

Ta có công thức :\(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\)

\(=\frac{1}{2}-\frac{1}{20}\)

\(=\frac{9}{20}\)

Ta có :

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)

\(=\frac{1}{2}-\frac{1}{20}=\frac{10}{20}-\frac{1}{20}=\frac{9}{20}\)

Vậy tổng \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}=\frac{9}{20}\)

Ta có : \(\frac{1}{1.2}\)+  \(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)

= 1  - \(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)

= 1 - \(\frac{1}{7}\)=  \(\frac{6}{7}\)

=1-1/2+1/2-1/3+1/3-1/4+...+1/6-1/7=1-1/7=6/7

1/1 -1/2 +1/2 -1/3 +1/3 -1/4 +1/4 -1/5 +1/5 -1/6 +1/6 -1/7

1/1- 1/7 =6/7

1/2.3+1/3.4+1/4.5+...+1/19.20

=1/2-1/3+1/3-1/4+1/4-1/5+...+1/19-1/20

=1/2-1/20

=9/20

HT NHA BN

3A=1.2.3+2.3.3+3.4.3+...+19.20.3

3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+19.20.(21-18)

3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+19.20.21-18.19.20

3A=19.20.21

=> \(A=\frac{19.20.21}{3}=2660\)

mk dùng cách của lớp 8 nha bạn ;

ta có công thức xích ma như sau x(x+1)

nhập vào xích ma ta có kết quả 2660

Ta có;

A=1.2+2.3+3.4+.....+19.20

3A=1.2.3+2.3.(4-1)+3.4.(5-2)+....+19.20.(21-18)

3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.....+19.20.21-18.19.20

A=19.20.21 = 2660

        3

\(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\cdot\cdot\cdot+\dfrac{1}{18\cdot19}+\dfrac{1}{19\cdot20}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\cdot\cdot\cdot+\dfrac{1}{18}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}\)

\(=\dfrac{1}{2}-\dfrac{1}{20}\)

\(=\dfrac{9}{20}\)

#\(Urushi\)☕

a) Ta có công thức tính tổng các số tự nhiên liên tiếp sau:

\(\Rightarrow1275=\frac{\left(1+n\right)n}{2}\Rightarrow\left(1+n\right)n=1275.2=2550=50.51\)

Mà n là số tự nhiên => n và n+1 là 2 số tự nhiên liên tiếp => n=50.

b) Đề chưa đầy đủ.

c) Ta có:

\(A=1.2+2.3+3.4+.....+19.20\)

\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+19.20.\left(21-18\right)\)

\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.....+19.20.21-18.19.20\)

\(=\left(1.2.3+2.3.4+3.4.5+......+19.20.21\right)-\left(1.2.3+2.3.4+......+18.19.20\right)=19.20.21\)

\(\Rightarrow A=19.20.7=2660=133.2.10\Rightarrow\frac{A}{133.2}=\frac{2.133.10}{2.133}=10\)

b) (n+1) . ( n+1) . (n+3) . ... . (n+100) + 7450

\(D=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)

\(=\frac{1}{2}-\frac{1}{20}\)

\(=\frac{9}{20}\)

\(D=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{19.20}\)

\(D=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{19}-\frac{1}{20}\)

\(D=\frac{1}{2}-\frac{1}{20}\)

\(D=\frac{9}{20}\)

Vậy : . . .

HOK TỐT