A=3(1/1.2+1/2.3+...+1/99.100)

A=3(1-1/2+1/2-1/3+...+1/99-1/100)

A=3(1-1/100)

A=3 . 99/100

A= 297 /100

5B= 1.2.3.4.5+2.3.4.5.5+....+97.98.99.100.5

   =1.2.3.4.5+2.3.4.5.6 -1.2.3.4.5+...+-96.97.98.99

=97.98.99.100.101=9505049400

=> B=1901009880

\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{97.98.99.100}=\frac{1}{3}.\left(\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...+\frac{3}{97.98.99.100}\right)=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{97.98.99}-\frac{1}{98.99.100}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{98.99.100}\right)=\frac{1}{3}.\left(\frac{1}{6}-\frac{1}{970200}\right)=\frac{1}{18}-\frac{1}{6.970200}\)

        \(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{97.98.99.100}\)

\(=\frac{1}{3}.\left(\frac{3}{1.2.3.4}+ \frac{3}{2.3.4.5}+...+\frac{3}{97.98.99.100}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{97.98.99}-\frac{1}{98.99.100}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{98.99.100}\right)\)

\(=\frac{1}{3}.\frac{161699}{970200}=\frac{161699}{299106000}\)

hai bạn trước đó gửi sai hết rùi. đúng theo sách NÂNG CAO PHÁT TRIỂN TOÁN 6 TẬP 2 thì bài này có đáp án thì bằng 1353/8120 nhé

ta có :

\(3A=-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)

\(\Rightarrow4A=-1-\frac{1}{3^{101}}\)

\(\Rightarrow4A=\frac{-3^{101}-1}{3^{101}}\)

\(\Rightarrow A=\left(\frac{-3^{101}-1}{3^{101}}\right):4\)

\(A=\frac{-1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\)

\(\Rightarrow3A=-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)

\(\Rightarrow3A+A=4A\)

 \(=\left(-1+\frac{1}{3}-...-\frac{1}{3^{100}}\right)+\left(\frac{-1}{3}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\right)\)

\(=-1+\frac{1}{3}-...-\frac{1}{3^{100}}-\frac{1}{3}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\)

\(=-1-\frac{1}{3^{101}}\)

\(\Rightarrow A=\frac{-1-\frac{1}{3^{101}}}{4}\)

Vậy \(A=\frac{-1-\frac{1}{3^{101}}}{4}\)

P=1/1.2.3.4 +1/2.3.4.5 +1/3.4.5.6 +...+1/97.98.99.100 

3P=3/1.2.3.4 +3/2.3.4.5 +3/3.4.5.6 +...+3/97.98.99.100

3P=1/1.2.3-1/2.3.4+1/2.3.4-1/3.4.5+................+1/97.98.99-1/98.99.100

3P = 1/1.2.3 - 1/98.99.100

3P =( 98.99.100-1.2.3)/1.2.3.98.99.100

P=( 98.99.100-1.2.3)/1.2.3.98.99.100.3

P=(98.33.50-1)/98.99.100.3

P= 161699/2910600

=398759

mk cũng cần lắm, ở trong toán vui mỗi tuần trên online math

cách làm của tui đúng nhất nhưng ko bít có giống cách ai ko

đặt S=1.2.3.4+2.3.4.5+3.4.5.6+...+97.98.99.100

5S=(5-0).1.2.3.4+(6-1).2.3.4.5+...+(101-96).97.98.99.100

5S=1.2.3.4.5-0+2.3.4.5.6-1.2.3.4.5+...+97.98.99.100.101-96.97.98.99.100

5S=97.98.99.100.101=9505049400

S=1901009880

1.2.3.4+2.3.4.5+3.4.5.6+...+97.98.99.100

4S=(1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100). 4

4S=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+98.99.100(101-97)

4S=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+4.5.6.7-3.4.5.6+...98.99.100.101-97.98.99.100

4S=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-3.4.5.6+...+97.98.99.100-97.98.99.100+98+99.100+101

4S=98.99.100.101

Vậy S = 98.99.100.101/4 = 24497550

1.2.3.4+2.3.4.5+3.4.5.6+...+97.98.99.100

4S=(1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100) 4

4S=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+98.99.100(101-97)

4S=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+4.5.6.7-3.4.5.6+...98.99.100.101-97.98.99.100

4S=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-3.4.5.6+...+97.98.99.100-97.98.99.100+98+99.100+101

4S=98.99.100.101

Vậy S=98.99.100.101/4

A=1/3+1/9+1/27+....+1/72171

3A=1/3x3+1/9x3+1/27x3+....+1/72171x3

3A=1+1/3+1/9+1/27+....+1/24057

3A-A=(1+1/3+1/9+1/27+...+1/24057)-(1/3+1/9+1/27+...+1/72171)

3A-A=1+1/3+1/9+1/27+...+72171-1/3-1/9-1/27-....-1/72171

=1-1/72171

2A=72170/72171

A=72170/72171:2

A=36085/72171

ta có: \(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{100^2}=1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)

Lại có: \(\frac{1}{2^2}>\frac{1}{2.3};\frac{1}{3^2}>\frac{1}{3.4};\frac{1}{4^2}>\frac{1}{4.5};...;\frac{1}{100^2}>\frac{1}{100.101}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\)

                                                                               \(=\frac{1}{2}-\frac{1}{101}\)

\(\Rightarrow1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)>1-\left(\frac{1}{2}-\frac{1}{101}\right)=1-\frac{1}{2}+\frac{1}{101}\)

                                                                                                                                 \(=\frac{1}{2}+\frac{1}{101}\)

mà \(\frac{1}{2}=\frac{50}{100}>\frac{1}{100}\Rightarrow\frac{1}{2}+\frac{1}{101}>\frac{1}{100}\)

=> đ p c m