Cho A = 1.2 + 2.3 + 3.4 + ... + 50.51 và B = 12 + 22 + 32 + ... + 492 + 502. Tính A - B
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\(a,A=1\cdot2+2\cdot3+...+98\cdot99\\ 3A=1\cdot2\cdot3+2\cdot3\cdot3+3\cdot4\cdot3+...+98\cdot99\cdot3\\ 3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\left(5-2\right)+...+98\cdot99\left(100-97\right)\\ 3A=1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4-2\cdot3\cdot4+3\cdot4\cdot5-...-97\cdot98\cdot99+98\cdot99\cdot100\\ 3A=98\cdot99\cdot100=970200\\ A=323400\)
\(b,B=1^2+2^2+3^3+...+98^2\\ B=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+98\left(99-1\right)\\ B=\left(1\cdot2+2\cdot3+3\cdot4+...+98\cdot99\right)-\left(1+2+...+98\right)\\ B=323400-\left[\left(98+1\right)\left(98-1+1\right):2\right]\\ B=323400-4851=318549\\ c,C=1\cdot99+2\left(99-1\right)+3\left(99-2\right)+...+98\left(99-97\right)+99\left(99-98\right)\\ C=1\cdot99+2\cdot99-1\cdot2+3\cdot99-2\cdot3+...+98\cdot99-97\cdot98+99\cdot99-98\cdot99\\ C=99\left(1+2+...+99\right)-\left(1\cdot2+2\cdot3+...+98\cdot99\right)\\ C=99\left[\left(99+1\right)\left(99-1+1\right):2\right]-323400\\ C=490050-323400=166650\)
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Lời giải:
$A=1.2+2.3+3.4+...+50.51$
$3A=1.2(3-0)+2.3(4-1)+3.4(5-2)+...+50.51(52-49)$
$=(1.2.3+2.3.4+3.4.5+...+50.51.52)-(0.1.2+1.2.3+2.3.4+....+49.50.51)$
$=50.51.52$
$\Rightarrow A=50.51.52:3=44200$
Ta có: 3S = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) + .....+ 50.51.(52 -49)
= 1.2.3 - 0 + 2.3.4 - 1.2.3 + 3.4.5 -2.3.4 + .....+ 50.51.52 - 49.50.51
3S = 50.51.52
S = 50.17.52 =44200
3S = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) + .....+ 50.51.(52 -49)
= 1.2.3 - 0 + 2.3.4 - 1.2.3 + 3.4.5 -2.3.4 + .....+ 50.51.52 - 49.50.51
3S = 50.51.52
S = 50.17.52 =44200
A=2(1-3)+4(5-3)+ 6(5-7)+...+50(49-57)
A=-4-8-12-...-100 = -(4+8+12+...+100) (tính tổng cấp số cộng)
Ta có:A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+......+\dfrac{1}{49.50}+\dfrac{1}{50.51}\)
A=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.......+\dfrac{1}{49}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{51}\)
A=1-\(\dfrac{1}{51}=\dfrac{50}{51}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}+\dfrac{1}{50.51}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{51}\)
\(A=\dfrac{1}{1}-\dfrac{1}{51}\)
\(A=\dfrac{50}{51}\)
\(A=1\cdot2+2\cdot3+...+50\cdot51\)
\(=1\cdot\left(1+1\right)+2\left(2+1\right)+3\left(3+1\right)+...+50\left(50+1\right)\)
\(=\left(1^2+2^2+...+50^2\right)+\left(1+2+...+51\right)\)
\(=B+\left(1+2+...+51\right)\)
\(=B+\dfrac{51\cdot52}{2}=B+51\cdot26\)
=>\(A-B=51\cdot26=1326\)
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