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D=5/4+5/4^2+5/4^3+....+5/4^99
4D=4(5/4+5/4^2+5/4^3+....+5/4^99)
4D=5+5/4+5/4^2+5/4^3+....+5/4^98
Lấy 4D-D=(5+5/4+5/4^2+5/4^3+....+5/4^98)-(5/4+5/4^2+5/4^3+....+5/4^99)
=>3D=5-5/4^99
D=5/3-5/3x4^99<5/3
=>d<5/3(ĐPCM)
`Answer:`
\(D=\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+\frac{5}{4^4}+...+\frac{5}{4^{99}}\)
\(\Rightarrow4D=5+\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{98}}\)
\(\Rightarrow4D-D=\left(5+\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{98}}\right)-\left(5+\frac{5}{4^2}+\frac{5}{4^3}+\frac{5}{4^4}+...+\frac{5}{4^{99}}\right)\)
\(\Rightarrow3D=5-\frac{5}{4^{99}}\)
\(\Rightarrow D=\left(5-\frac{5}{4^{99}}\right):3\)
\(\Rightarrow D=\frac{5}{3}-\frac{5}{4^{99}.3}< \frac{5}{3}\)
Vậy `D<5/3`
\(D=4+4^2+4^3+4^4+...+4^{100}\)
\(4D=4^2+4^3+4^4+4^5+...+4^{101}\)
\(4D-D=\left(4^2+4^3+4^4+4^5+...+4^{101}\right)-\left(4+4^2+4^3+4^4+...+4^{100}\right)\)
\(3D=4^{101}-4\)
\(D=\dfrac{4^{101}-4}{3}\)
\(#WendyDang\)
\(D=4+4^2+4^3+...+4^{100}\)
=>\(4D=4^2+4^3+...+4^{101}\)
=>\(4D-D=4^{101}+4^{100}+4^{99}+...+4^3+4^2-4^{100}-4^{99}-...-4^2-4\)
=>\(3D=4^{101}-4\)
=>\(D=\dfrac{4^{101}-4}{3}\)
`#3107.101107`
\(D=4 + 4^2 + 4^3 + 4^4 + …. + 4^{100}\)
\(4D=4^2 + 4^3 + 4^4 + ... + 4^{101}\)
\(4D - D = (4^2 + 4^3 + 4^4 ... + 4^{101}) - (4 + 4^2 + 4^3 + ... + 4^{100})\)
\(3D = 4^2 + 4^3 + 4^4 + ... + 4^{101} - 4 - 4^2 - 4^3 - ... - 4^{100}\)
\(3D = 4^{101} - 4\)
\(D = \dfrac{4^{101} - 4}{3}\)
Vậy, \(D=\dfrac{4^{101} - 4}{3}.\)
Ta có: \(D=1+4^2+4^4+4^6+...+4^{100}\)
=> \(16D=4^2+4^4+4^6+4^8+...+4^{102}\)
=> \(16D-D=\left(4^2+4^4+...+4^{102}\right)-\left(1+4^2+...+4^{100}\right)\)
<=> \(15D=4^{102}-1\)
=> \(D=\frac{4^{102}-1}{15}\)
\(D=4-4^2+4^3-4^4+...+4^{2024}\\ 4D=4^2-4^3+4^4-4^5+...+4^{2025}\\ 4D+D=\left(4^2-4^3+4^4-4^5+...+4^{2025}\right)+\left(4-4^2+4^3-4^4+...+4^{2024}\right)\\ 5D=4^{2025}+4\\ D=\dfrac{4^{2025}+4}{5}\)