\(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+...+\frac{4}{99\cdot101}-x-\frac{200}{101}=1\)

\(\frac{4}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)-x=1+\frac{200}{101}\)

\(\frac{4}{2}\cdot\left(1-\frac{1}{101}\right)-x=\frac{301}{101}\)

\(\frac{4}{2}\cdot\frac{100}{101}-x=\frac{301}{101}\)

\(\frac{200}{101}-x=\frac{301}{101}\)

\(\Rightarrow x=\frac{301}{101}-\frac{200}{101}=1\)

Ta có : \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+.....+\frac{4}{99.101}-x-\frac{200}{101}=1\)

\(\Rightarrow\)\(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+.....+\frac{4}{99.101}=1+\frac{200}{101}+x\)

=> \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+.....+\frac{4}{99.101}=\frac{301}{101}+x\)

=> \(2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.101}\right)=\frac{301}{101}+x\)

=> \(2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{99}-\frac{1}{101}\right)=\frac{301}{101}+x\)

=> \(2\left(1-\frac{1}{101}\right)=\frac{301}{101}+x\)

=> \(2.\frac{100}{101}=\frac{301}{101}+x\)

=> \(\frac{200}{101}=\frac{301}{101}+x\)

\(\Rightarrow x=\frac{301}{101}-\frac{200}{101}=1\)

Theo công thức là ra nhé=))

chịu

A=1x3x(5+1) + 3x5x(7-1) +5x7x(9-3) +...+ 99x101x(103-97)

6A=3+ 1x3x5 +3x5x7-1x3x5 + 5x7x9 -3x5x7 +....+99x101x103 - 97x99x101

6A=3+99x101x103=1019703

vậy = 1019703

nếu sai chỗ nào thì sửa hộ mk vs

=1(2+1)+2(3+1)+3(4+1)+...+100(101+1)

=1.2+1+2.3+2+3.4+3+...+100.101+100

=(1.2+2.3+3.4+..+100.101)+(1+2+3+...+100)

=333300+5000

=338300

A=1x3 +3x5 +5x7 +....+99x101

6A=1x3x(5+1) + 3x5x(7-1) +5x7x(9-3) +...+ 99x101x(103-97)

6A=3+ 1x3x5 +3x5x7-1x3x5 + 5x7x9 -3x5x7 +....+99x101x103 - 97x99x101

6A=3+99x101x103=1019703

\(S=1.3+2.4+3.5+...+99.101\)

\(\Rightarrow S=1\left(2+1\right)+2\left(3+1\right)+...+99\left(100+1\right)\)

\(\Rightarrow S=\left(1.2+2.3+...+99.100\right)+\left(1+2+3+...+99\right)\)

Đặt \(A=1.2+2.3+...+99.100\)

\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+...+99.100.\left(101-98\right)\)

\(\Rightarrow3S=1.2.3+2.3.4-1.2.3+...+99.100.101-98.99.100\)

\(\Rightarrow S=\frac{99.100.101}{3}\)

Đặt \(B=1+2+3+...+99\)

\(\Rightarrow B=\frac{\left(99+1\right)\left[\left(99-1\right):2+1\right]}{2}\)

\(\Rightarrow B=\frac{100.50}{2}=2500\)

\(\Rightarrow S=A+B=\frac{99.100.101}{3}+2500\)

S = 1 x 3 + 2 x 4 + 3 x 5 + ... + 99 x 101

S = ( 1 x 3 + 3 x 5 + ...+ 99 x 101) +  ( 2 x 4 + ...+ 98 x 100)

Đặt A = 1 x 3 + 3 x 5 + ...+ 99 x 101

=> 6 A = 1 x 3 x 6 + 3 x 5 x 6 + ...+ 99 x 101 x 6

6 A = 1 x 3 x ( 5+1) + 3 x 5 x ( 7-1) + ...+ 99 x 101 x ( 103 - 97)

6A = 1 x 3 x 5 + 1 x 3 + 3 x 5 x 7 - 1 x 3 x 5 + ...+ 99 x 101 x 103 - 97 x 99 x 101

6A = ( 1 x 3 + 1 x 3 x 5 + 3 x 5 x 7 +...+ 99 x 101 x 103) - ( 1 x 3 x 5 + ...+ 97 x 99 x 101)

6A = 1 x  3 + 99 x 101 x 103

\(\Rightarrow A=\frac{1.3+99.101.103}{6}=171650\)

Đặt B = 2 x 4 + ...+ 98 x 100

=> 6B = 2 x 4 x 6 + 4 x 6 x 6 + ...+ 98 x 100 x 6

6B = 2 x 4 x 6 + 4 x 6 x ( 8-2) + ...+ 98 x 100 x ( 102 - 96)

6B = 2 x 4 x 6 + 4 x6 x8 - 2x4x6 + ...+ 98x100x102 - 96x98x100

6B = ( 2 x 4 x 6 + 4 x 6 x 8 +...+98x100x102) - ( 2x4x6+...+96x98x100)

6B = 98 x 100 x 102

\(\Rightarrow B=\frac{98.100.102}{6}=166600\)

Thay A;B vào S, có
S = 171 650 + 166 600

S = 338 250