1x2+2x3+3x4+..+98x99
Tính tổng: 1x2+2x3+3x4+...+98x99
A= 1x2+2x3+3x4+...+98x99 A x 3= 1x2 x (3-0) +2x3x (4-1)+3x4 x (5-2)+...+98x99x (100-97) = 1x2x3+2x3x4+......98x99x100- (1x2x0+ 2x3x1+....+ 98x99x97) = 98x99x100
Tính tổng: 1x2+2x3+3x4+...+98x99
A= 1x2+2x3+3x4+...+98x99
A x 3= 1x2 x (3-0) +2x3x (4-1)+3x4 x (5-2)+...+98x99x (100-97)
= 1x2x3+2x3x4+......98x99x100- (1x2x0+ 2x3x1+....+ 98x99x97)
= 98x99x100.
1x2+2x3+3x4+...+98x99+99x100 = ?
Đặt S = 1x2+2x3+3x4+...+98x99+99x100
S x 3 =1x2x3+2x3x3+3x4x3+...+98x99x3+99x100x3
S x 3 =1x2x(3-0)+2x3x(4-1)+3x4x(5-2)+....+98x99x(100-97)+99x100x(101-98)
S x 3 = 1x2x3 + 2x3x4-1x2x3+3x4x5-2x3x4+...+98x99x100-97x98x99+99x100x101-98x99x100
S x 3 = 99x100x101
S x 3 = 999900
S = 333300
S= 1x2+2x3+3x4+4x5+...+98x99+99x100
Gọi biểu thức trên là A, ta có :
A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
A x 3 = 99x100x101
A = 99x100x101 : 3
A = 333300
3/1x2-5/2x3+7/3x4-9/4x5+.......-197/98x99
9/1x2+9/2x3+9/3x4+.................+9/98x99+9/99x100
Cho tổng trên là A
Ta co :
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(A=9.\frac{99}{100}\)
\(A=\frac{891}{100}\)
1/1x2 + 1/2x3 + 1/3x4 +......+1/98x99 + 1/99x100
1/1.2 +1/2.3 +1/3.4 +...+1/98.99 +1/99.100
=1-1/2+1/2-1/3+1/3-1/4+...+1/98-1/99+1/99-1/100
=1-1/100=100/100-1/100=99/100
Ta có: \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow1-\frac{1}{100}=\frac{99}{100}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
= \(1-\frac{1}{100}=\frac{99}{100}\)
S=1x2+2x3+3x4+4x5+...+98x99
Tìm S:))
S=1x2+2x3+3x4+4x5+...+98x99
3S= 1.2.3+ 2.3.3 + 3.4.3 + 4.5.3+...+98.99.3
3S= 1.2.3+ 2.3(4-1) + 3.4(5-2) + 4.5(6-3)+....+ 98.99.(100-97)
3S= 1.2.3 + 2.3.4 -1.2.3 + 3.4.5 - 2.3.4 +...+98.99.100 -97.98.99
3S= 98.99.100
S=970200:3
S= 323400
Bài làm:
\(S=1.2+2.3+3.4+...+98.99\)
\(S=\frac{1}{3}\left(1.2.3+2.3.3+3.4.3+...+98.99.3\right)\)
\(S=\frac{1}{3}\left[1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+98.99.\left(100-97\right)\right]\)
\(S=\frac{1}{3}\left(1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-97.98.99+98.99.100\right)\)
\(S=\frac{98.99.100}{3}=323400\)
Vậy S = 323400
Học tốt!!!!
Tính:1/1x2 + 1/2x3 + 1/3x4 + ... + 1/98x99 + 1/99x100
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Cho hai số biết rằng bớt số thứ nhất 28 đơn vị thì được số thứ hai va 1/3 số thứ nhất bằng 3/5 số thứ hai.Tìm hai số đó