\(A=1+2+2^2+...+2^{2018}\)
\(2A=2+2^2+...+2^{2019}\)
\(2A-A=\left[2+2^2+...+2^{2019}\right]-\left[1+2+2^2+...+2^{2018}\right]\)
\(A=2^{2019}-1\)
#)Giải :
\(A=1+2+2^2+2^3+...+2^{2018}\)
\(2A=2+2^2+2^3+2^4+...+2^{2019}\)
\(2A-A=\left(2+2^2+2^3+2^4+...+2^{2019}\right)-\left(1+2+2^2+2^3+...+2^{2018}\right)\)
\(A=2^{2019}-1\)
\(B=3+3^2+3^3+...+3^{2017}\)
\(3B=3^2+3^3+3^4+...+3^{2018}\)
\(3B-B=\left(3^2+3^3+3^4+...+3^{2018}\right)-\left(3+3^2+3^3+...+3^{2017}\right)\)
\(2B=3^{2018}-3\)
\(B=\frac{3^{2018}-3}{2}\)
\(C=1+5^2+5^4+...+5^{2018}\)
\(5^2C=5^2+5^4+...+5^{2020}\)
\(5^2C-C=\left[5^2+5^4+...+5^{2020}\right]-\left[1+5^2+5^4+...+5^{2018}\right]\)
\(24C=5^{2020}-1\)
\(C=\frac{5^{2020}-1}{24}\)
D = 1.3 + 2.4 + 3.5 + ... + 99.101
D = 1.(2 + 1) + 2.(3 + 1) + 3.(4 + 1) + .... + 99.(100 + 1)
D = (1.2 + 2.3 + 3.4 + ... + 99.100) + (1 + 2 + 3 + ... + 99 )
Đặt S = 1.2 + 2.3 + 3.4 + ... + 99.100
3S = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3
3S = 1.2.3 + 2.3.(4-1) + 3.4.(5 -2) + ... + 99.100.(101-98)
3S = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100
3S = (1.2.3 + 2.3.4 + 3.4.5 + ... + 99.100.101) - (1.2.3 + 2.3.4 + ... + 98.99.100)
3S = 99.100.101
S = \(\frac{99.100.101}{3}=\frac{999900}{3}=333300\)
Thay S vào biểu thức D , ta có :
\(333300+\left(1+2+...+99\right)\)
\(=333300+\frac{\left(99+1\right)99}{2}\)
\(=333300+4950=338250\)
Vậy D = 338250
