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\(1^2-2^2+3^2-4^2+...+97^2-98^2+99^2-100^2=\left(1-2\right)\left(1+2\right)+\left(3-4\right)\left(3+4\right)+...+\left(97-98\right)\left(97+98\right)+\left(99-100\right)\left(99+100\right)\)\(=-\left(1+2+3+4+...+97+98+99+100\right)\)
\(=-\left(\frac{101\times100}{2}\right)=-5050\)
![](https://rs.olm.vn/images/avt/0.png?1311)
S = 1.2.3.4 + 2.3.4.5 + 3.4.5.6+...97.98.99.100
5S = (1.2.3.4+2.3.4.5+3.4.5.6+ ... + 97.98.99.100).5
5S = 1.2.3.4.(5-0) + 2.3.4.5.(6-1)+ 3.4.5.6(7-2)+......+ 97.98.99.100.(101-96)
5S = (1.2.3.4.5 + 2.3.4.5.6 + 3.4.5.6.7 + ....+ 97.98.99.100.101) - (0.1.2.3.4 + 1.2.3.4.5 + 2.3.4.5.6+.....+96.97.98.99.100)
5S = 97.98.99.100.101
S= 97.98.99.100.101/5
S=1901009880
S=1*2*3*4+2*3*4*5+....+97*98*99*100
5S=1.2.3.4.5+2.3.4.5.5+...+97.98.99.100.5
5S=1.2.3.4.(5-0)+2.3.4.5.(6-1)+...+97.98.99.100.(101-96)
5S=1.2.3.4.5-0.1.2.3.4+2.3.4.5.6-1.2.3.4.5+...+97.98.99.100.101-96.97.98.99.100
5S=(1.2.3.4.5+2.3.4.5.6+...+97.98.99.100.101)-(0.1.2.3.4+1.2.3.4.5+...+96.97.98.99.100)
5S=97.98.99.100.101
S=9505049400:5=1901009880.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(100^2+98^2+...+2^2\right)-\left(99^2+97^2+...+1^2\right)\)
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+....+\left(2^2-1^2\right)\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=100+99+98+97+....+2+1=5050\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(100^2+98^2+...+2^2\right)-\left(99^2+97^2+...+1^2\right)\)
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+....+\left(2^2-1^2\right)\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=100+99+98+97+....+2+1=5050\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=100+99+98+97+...+2+1\)
\(=101\cdot50=5050\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=100^2-99^2+98^2-97^2+....+2^2-1^2\)
\(=\left(100-99\right).\left(100+99\right)+\left(98-97\right).\left(98+97\right)+....+\left(2-1\right).\left(2+1\right)\)
\(=1+2+....+97+98+99+100=\frac{100.\left(100+1\right)}{2}=5050\)
\(B=3\left(2^2+1\right)\left(2^4+1\right)....\left(2^{64}+1\right)+1=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(=\left(2^4-1\right)\left(2^4+1\right)......\left(2^{64}+1\right)+1=\left(2^8-1\right).....\left(2^{64}+1\right)+1\)
Tiếp tục rút gọn như vậy,ta đc \(B=\left(2^{64}-1\right)\left(2^{64}+1\right)=2^{128}-1+1=2^{128}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(S=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(S=\left(100-99\right)\left(99+100\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(S=1\cdot199+1\cdot195+1\cdot193+...+1\cdot3\)
\(S=199+195+193+...+3\)
\(S=\left(3+199\right)\cdot\left[\left(199-3\right):2+1\right]:2\)
\(S=202\cdot99:2\)
\(S=101\cdot99\)
\(S=9999\)
![](https://rs.olm.vn/images/avt/0.png?1311)
100^2-99^2 = (100-99)x(100+99) =199
tương tự 98^2-97^2=195
.....................
2^2 - 1^2 = 3
=> cái bỉu thức trên thành
199+195+191+....+3
số số hạng: (199-3):4 + 1=50
tổng: [(199+3)*50]:2=5050
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= (100^2 - 99^2) + (98^2 - 97^2) + ... + (4^2 - 3^2) + (2^2 - 1^2) =
= (100+99)(100-99) + (98+97)(98-97) + ... + (4+3)(4-3) + (2+1)(2-1) =
= (100+99).1 + (98+97).1 + ... + (4+3).1 + (2+1).1 =
= 100 + 99 + 98 + 97 + ... + 4 + 3 + 2 + 1 =
= (100+1) + (99+2) + (98+3) + ... + (51+50) = 101.50 = 5050
(50 cặp dấu ngoặc, tổng trong mỗi cặp dấu ngoặc là 101)