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\(A=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)
\(3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+99\cdot100\cdot\left(101-98\right)\)
\(3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+...+99\cdot100\cdot101-98\cdot99\cdot100\)
\(3A=99\cdot100\cdot101\Rightarrow A=\dfrac{99\cdot100\cdot101}{3}=333300\)
\(B=1^2+2^2+3^2+...+99^2+100^2\)
\(=\dfrac{100\cdot\left(100+1\right)\cdot\left(2\cdot100+1\right)}{6}\)
\(=\dfrac{2030100}{6}=338350\)
\(C=1\cdot2\cdot3+2\cdot3\cdot4+...+8\cdot9\cdot10\)
\(4C=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot\left(5-1\right)+...+8\cdot9\cdot10\cdot\left(11-7\right)\)
\(4C=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+...+8\cdot9\cdot10\cdot11-7\cdot8\cdot9\cdot10\)
\(4C=8\cdot9\cdot10\cdot11\Rightarrow C=\dfrac{8\cdot9\cdot10\cdot11}{4}=1980\)
MÌNH KO GHI LẠI ĐỀ NHA
A=100.(100-1).(100+1):3
A=333300
B=100.(100+1).(100.2+1):6
B=100.101.201:6
B=338350
\(A=4+4^2+4^3+....+4^{99}+4^{100}\)
\(=4\left(4+1\right)+4^3\left(4+1\right)+...+4^{99}\left(4+1\right)\)
\(=4\cdot5+4^3\cdot5+...+4^{99}\cdot5\)
\(=5\left(4+4^3+...+4^{99}\right)\)
\(S=1\cdot2+2\cdot3+3\cdot4+...+2018\cdot2019\)
\(3S=1\cdot2\cdot3+2\cdot3\cdot3+3\cdot3\cdot4+...+2018\cdot2019\cdot3\)
\(3S=1\cdot2\cdot\left(3-0\right)+2\cdot3\left(4-1\right)+....+2018\cdot2019\left(2020-2017\right)\)
\(3S=1\cdot2\cdot3-0\cdot1\cdot2+2\cdot3\cdot4-1\cdot2\cdot3+....+2018\cdot2019\cdot2020-2017\cdot2018\cdot2019\)
\(3S=2018\cdot2019\cdot2020\)
\(S=\frac{2018\cdot2019\cdot2020}{3}\)
\(1\cdot2\cdot3+2\cdot3\cdot4+...+48\cdot49\cdot50\)
\(4P=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+...+48\cdot49\cdot50\cdot4\)
\(4P=1\cdot2\cdot3\left(4-0\right)+2\cdot3\cdot4\left(5-1\right)+....+48\cdot49\cdot50\left(51-47\right)\)
\(4P=1\cdot2\cdot3\cdot4-0\cdot1\cdot2\cdot3+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+....+48\cdot49\cdot50\cdot51-47\cdot48\cdot49\cdot50\)
\(P=\frac{48\cdot49\cdot50\cdot51}{4}\)
\(Q=1^2+2^2+3^2+....+113^2\)
\(Q=1\left(2-1\right)+2\left(3-1\right)+....+133\left(134-1\right)\)
\(Q=\left(1\cdot2+2\cdot3+133\cdot134\right)-\left(1+2+3+...+133\right)\)
Áp dụng công thức cho nó nhanh:\(1\cdot2+2\cdot3+...+133\cdot134=\frac{133\cdot134\cdot135}{3}\)
\(1+2+3+...+133=\frac{133\cdot134}{2}\)
Đến đây đưa casio ra mak tính
c, 4C= (1.2.3+2.3.4+3.4.5+...+8.9.10) .4
==> 4C= [1.2.3.(4-0) + 2.3.4-(5-1) + 8.9.10.(11-7)
==>4C= 1.2.3.4 - 1.2.3.4+ 2.3.4.5-2.3.4.5 + 7.8.9.10- 7.8.9.10 + 8.9.10.11
==> 4C= 8.9.10.11=7920
==> C= 7920 :4=1980
a, Ta có: 3A= 1.2.3+2.3.3+3.4.3+...+99.100.3
3A=1.2.(3-0) + 2.3.(4-1)+ 3.4.(5-2)+ ... + 99.100.( 101-98)
3A=(1.2.3 + 2.3.4+ 3.4.5+ 99.100.101) - (0.1.2 +1.2.3+ 2.3.4 + ... + 98.99.100)
3A= 99.100.101 - 0.1.2
3A= 999900 - 0
3A= 999900
==> A= 999900 : 3
==> A= 333300