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b1. 456 = 10.(40+5)+6
A = 10( 44.....440 + 55....55) + 66..66 (... 111 số)
=499.....9950 + 66...66 (... 111 số 9 và 111 số 6)
= 55....5516 (....111 số 5)
b2. A - B = 1+2 + 3 + 4 +....+98 = 49 x100 + 51 = 4951
\(b=1.1+2.2+...+98.98=1\left(2-1\right)+2\left(3-1\right)+..+98.\left(99-1\right)=\left(1.2+2.3+...+98.99\right)-\left(1+2+...+98\right)\)=> \(a-b=\left(1.2+2.3+..+98.99\right)-\left[\left(1.2+2.3+...+98.99\right)-\left(1+2+...+98\right)\right]=1+2+3+...+98\)ta tính tổng của dãy số: a-b= (98+1).98:2=4851
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
B = ... (bạn tự tính)
=> A - B = ...
A=1x2+2x3+3x4+...+49x50
3A= 3(1.2+2.3+3.4+...+49.50)
3A= 1.2.3+2.3.3+3.4.3+...+49.50.3
3A= 1.2.(3-0)+2.3(4-1)+3.4(5-2)+...+49.50.(51-48)
3A= 0.1.2-1.2.3+1.2.3-2.3.4+2.3.4-3.4.5+...+48.49.50-49.50.51
3A= 49.50.51
A= 49.50.51/3=41650
B=1x3+3x5+5x7+...+99x101
B=1/1.3 +1/3.5 +...+1/99.101
2B=2/1.3 + 2/3.5 +...+2/99.101
2B=1-1/3+1/3-1/5+...+1/99-1/101
2B=1-1/101
2B=100/101
B=100/101:2=100/202
S5=5x5-(4x4-(3x3-(2x2-1x1)))
S2011=2001x2001-(2000x2000-(1999x1999-(....)))
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
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+....+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9\left(1-\frac{1}{100}\right)\)
\(A=9\cdot\frac{99}{100}=\frac{891}{100}\)
\(A=9\left(\frac{1}{1x2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
=> \(A=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
=> \(A=9\left(1-\frac{1}{100}\right)=\frac{9.99}{100}=\frac{891}{100}\)
=> A=8,91
Ta có:\(A=\frac{9}{1.2}+\frac{9}{2.3}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(=9\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=9\left(1-\frac{1}{100}\right)\)
\(=9.\frac{99}{100}=\frac{891}{100}\)