Đặt A = \(3^1+3^2+3^3+3^4+.....+3^{100}\)
=>3A=3(\(3^1+3^2+3^3+.....+3^{100}\))
=>3A = \(3^2+3^3+3^4+....+3^{101}\)
=>3A-A = (\(3^2+3^3+3^4+.....+3^{101}\))-(\(3^1+3^2+3^3+....+3^{100}\))
=>2A= \(3^{101}-3^1\)
=>A=\(\frac{3^{101}-3}{2}\)
Đặt A = \(3^1+3^2+3^3+3^4+.....+3^{100}\)
=>3A=3(\(3^1+3^2+3^3+.....+3^{100}\))
=>3A = \(3^2+3^3+3^4+....+3^{101}\)
=>3A-A = (\(3^2+3^3+3^4+.....+3^{101}\))-(\(3^1+3^2+3^3+....+3^{100}\))
=>2A= \(3^{101}-3^1\)
=>A=\(\frac{3^{101}-3}{2}\)
A= 1/3 - 2/ 32 + 3/ 33 - 4/ 34 + .... + 99/ 399 - 100/ 3100 < 3/ 16
A= 1/3 - 2/ 32 + 3/ 33 - 4/ 34 + .... + 99/ 399 - 100/ 3100 < 3/ 16
thực hiện phép tính.
a) b) C= 3+31+32+33+...+3100 c)\(\dfrac{2018.2019-1}{2018^2+2017}\)
\(\dfrac{2^{10}.13+2^{10}.65}{2^8.104}\)
A = 3 + 32 + 33 + .... + 3100
Cho S=1/31+1/32+1/33+...+1/60 Chứng minh S<4/5("/" là phần)
Cho \(S=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{60}\) .
Chứng minh rằng \(3< 5S< 4\)
Cho S=1/31+1/32+1/33+...+1/60
CMR:3/5<S<4/5
cho S = 1/31+1/32+1/33+...+1/59+1/60. cmr 3/4<S<4/5
A = 1 phần 31 + 1 phần 32 + 1 phần 33 + .... + 1 phần 60
Chứng minh 3 phần 5 < A < 4 phần 5A=1/31+1/32+...+1/2048 > 3