Tính tổng: A=2^2 + 3^2 + 4^2 + ... + 100^2
A=2^2 + 3^2 + 4^2 + ... + 100^2
A=(2.3.4...100)+(2.3.4...100)+1
A=2(2.3.4....100)+1
A= ...
Tính tiếp, ngang đó mik chịu
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5,Ta có
A=1/2+1/2^2+1/2^3+...+1/2^100
2A=1+1/2+1/2^2+1^2/3+...+1/2^99
2A-A=(1+1/2+1/2^2+1^2/3+...+1/2^99)-(1/2+1/2^2+1/2^3+...+1/2^100)
A=1-1/2^100
Ta có : A = 1.2.3 + 2.3.4 + 4.5.6 + ..... + 98.99.100
=> 6A = 1.2.3.4 - 1.2.3.4 + 2.3.4.5 - 2.3.4.5 + ...... + 98.99.100.101
=> 6A = 98.99.100.101
=> A = \(\frac{98.99.100.101}{6}=16331700\)
có 20172 đồng dư 1 mod (3)
=> (20172)50 đồng dư 1 mod (3)
=> (20172)50-1 đồng dư 1-1 = 0 mod (3)
=> dpcm
Câu a)
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(=\left(2^{100}+2^{99}+2^{98}+2^{97}+...+2^2+2\right)-2\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)\)
\(=\left(2^{100}+2^{99}+2^{98}+2^{97}+...+2^2+2\right)-\left(2^{100}+2^{98}+2^{96}+...+2^4+2^2\right)\)
\(=2^{99}+2^{97}+2^{95}+...+2^3+2\)
\(=\frac{2^2\cdot\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)-\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)}{3}\)
\(=\frac{\left(2^{101}+2^{99}+2^{97}+...+2^5+2^3\right)-\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)}{3}\)
\(=\frac{2^{101}-2}{3}\)
\(2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2015.2016.2017}\)
\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{2.4}+...+\frac{1}{2015.2016}-\frac{1}{2016.2017}\)
\(2B=\frac{1}{1.2}-\frac{1}{2016.2017}\)
\(B=\frac{\frac{1}{1.2}-\frac{1}{2016.1017}}{2}\)
Tính
a,1.2.3+2.3.4+3.4.5+......+ 98.99.100
b,1 bình +2 bình +3 bình +....+100 bình
Giải:Đặt A=1.2.3+2.3.4+..........+98.99.100
4A=1.2.3.4+2.3.4.5-1.2.3.4+...........+98.99.100.101-97.98.99.100
4A=98.99.100.101=97990200\(\Rightarrow A=24497550\)
b,Đặt B=12+22+................+1002
B=1.(2-1)+2.(3-1)+.............+100.(101-1)
B=1.2+2.3+.......+100.101-1-2-..........-100
Đặt C=1.2+2.3+........+100.101
3C=1.2.3+2.3.4-1.2.3+........+100.101.102-99.100.101
3C=100.101.102=1030200\(\Rightarrow C=343400\)
\(\Rightarrow B=343400-\frac{100.101}{2}=343400-5050=338350\)