1x2+2x3+.........x99+100
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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
Ta có:
\(A=1.2+2.3+3.4+...+99.100\)
\(\Rightarrow3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)
\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)
\(\Leftrightarrow3A=99.100.101\Leftrightarrow A=\frac{99.100.101}{3}=333300\)
\(B=1.2.3+2.3.4+4.5.6+...+98.99.100\)
\(\Rightarrow4B=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+4.5.6.\left(7-3\right)+...+98.99.100.\left(101-97\right)\)
\(\Rightarrow4B=1.2.3.4+2.3.4.5-1.2.3.4+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100\)
\(\Leftrightarrow4B=98.99.100.101\Leftrightarrow B=\frac{98.99.100.101}{4}=24497550\)
#)Giải :
Đặt \(A=1.2+2.3+3.4+...+99.100\)
\(3A=1.2.3+2.3.3+3.4.3+...+49.50.3\)
\(3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)
\(3A=0.1.2-1.2.3+1.2.3-2.3.4+2.3.4-3.4.5+...+48.49.50-49.50.51\)
\(3A=49.50.51=124950\)
\(\Leftrightarrow A=\frac{124950}{3}=41650\)
Mình sửa lại đề vì sai : 1 x 2 + 2 x 3 + ... + 99 x 100
Đặt A = 1 x 2 + 2 x 3 + ... + 99 x 100
=> 3 x A = 3 x (1 x 2 + 2 x 3 + ... + 99 x 100)
=> 3 x A = 1 x 2 x 3 + 2 x 3 x 3 + ... + 99 x 100 x 3
=> 3 x A = 1 x 2 x 3 + 2 x 3 x (4 - 1) + ... + 99 x 100 x (101 - 98)
=> 3 x A = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + ... + 99 x 100 x 101 - 98 x 99 x 100
=> 3 x A = 99 x 100 x 101
=> 3 x A = 999 900
=> A = 999 900 : 3
=> A = 333 300
Vậy 1 x 2 + 2 x 3 + ... + 99 x 100 = 333 300