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B = 1(2-1) + 2(3-1) + 3(4-1) + ...+ 98(99-1)+99(100-1)=
=1.2+2.3+3.4+...+98.99+99.100-(1+2+3+...+99)
Đặt C=1.2+2.3+3.4+...+98.99+99.100
3C=1.2.3+2.3.3+3.4.3+...+98.99.3+99.100.3=1.2.3+2.3.(4-1)+3.4.(5-2)+...+98.99.(100-97)+99.100.(101-98)=
=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-97.98.99+98.99.100-98.99.100+99.100.101=99.100.101
\(\Rightarrow C=\frac{99.100.101}{3}=33.100.101\)
Đặt D=1+2+3+...+99 =\(\frac{99\left(1+99\right)}{2}=50.99\)
\(B=C-D=33.100.101-50.99=\)
A=\(1+2^1+2^2+2^3+2^4+...+2^{2015}\)
2A=\(2^1+2^2+2^3+2^4+...+2^{2015}+2^{2016}\)
2A-A=\(2^{2016}-1\)
Vậy A=\(2^{2016}-1\)
Đặt \(A=1+2+...+2^{97}+2^{98}+2^{99}\)\(\Rightarrow\)\(2^{100}-A=2^{100}-\left(1+2+...+2^{97}+2^{98}+2^{99}\right)\)
Ta có: \(2A=2+2^2...+2^{98}+2^{99}+2^{100}\)
Lấy \(2A-A\)theo vế, ta có:
\(2A-A=\left(2+2^2...+2^{98}+2^{99}+2^{100}\right)-\left(1+2+...+2^{97}+2^{98}+2^{99}\right)\)
\(\Leftrightarrow2A-A=2+2^2...+2^{98}+2^{99}+2^{100}-1-2-...-2^{97}-2^{98}-2^{99}\)
\(\Leftrightarrow A=2^{100}-1\)
\(\Rightarrow2^{100}-A=2^{100}-2^{100}+1=1\)
Vậy \(2^{100}-\left(1+2+...+2^{97}+2^{98}+2^{99}\right)=1\)
tinh tong cua 11,10,9,8,7,6,5,..........,-37,-38
-675 nha ban
\(\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
\(=\frac{11.3^{22}.3^7-\left(3^2\right)^{15}}{\left(2.3^{14}\right)^2}\)
\(=\frac{11.3^{22}.3^7-3^{30}}{\left(2.3^{14}\right)^2}\)
\(=\frac{11.3^{29}-3^{30}}{\left(2.3^{14}\right)^2}\)
\(=\frac{11.3^{29}-3^{30}}{2^2.3^{28}}\)
\(=\frac{11.3^{29}-3.3^{29}}{4.3^{28}}\)
\(=\frac{\left(11-3\right).3^{29}}{4.3^{28}}\)
\(=\frac{8.3^{29}}{4.3^{28}}\)
\(=2.3=6\)
=2040
sai rồi bn ơi