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\(A=3+3^2+3^3+...+3^{100}\)
\(\Leftrightarrow3A=3^2+3^3+3^4+3^5+....+3^{101}\)
\(\Leftrightarrow3A-A=\left(3^2+3^3+3^4+3^5+...+3^{101}\right)-\left(3+3^2+3^3+3^4+...+3^{100}\right)\)
\(\Leftrightarrow2A=3^{101}-3\)
\(\Leftrightarrow A=\frac{3^{101}-3}{2}< 3^{100}-1\)
\(\Leftrightarrow A< B\)
a. tính A = 3+3^2+3^3+3^4+.....+3^100
3A=3^2+3^3+3^4+3^5+....+3^100
3A-A=(3^2+3^3+3^4+....+3^101)-(3+3^2+3^3+3^4+.....+3^100)=3^101-3=3^100
mà B=3^100-1 => A<B
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\(\Rightarrow3B=3^2+3^3+3^4+...+3^{101}\\ \Rightarrow3B-B=3^2+3^3+...+3^{101}-3-3^2-3^3-...-3^{100}\\ \Rightarrow2B=3^{101}-3\\ \Rightarrow B=\dfrac{3^{101}-3}{2}\)
B = 31 + 32 + 33 + .... + 399 + 3100
3B = 3(31 + 32 + 33 + ..... + 399 + 3100)
3B = 32 + 33 + 34 +...... + 3100 + 3101
3B - B = 2B = (32 + 33 + 34 + .... + 3100 + 3101) - ( 31 + 32 + 33 + .... + 3100)
2B = (32 - 32) + (33 - 33) +.....+ ( 3100 - 3100) + ( 3101 - 1)
2B = 0 + 0 + 0 + ..... +0 + 3101 - 1
2B = 3101 - 1
B = (3101 - 1) : 2
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A=1+4+42+43+.......+499 4A=4+42+43+44+.....+4100 4A-A=4+42+43+44+.....+4100 -1-4-42-43-.......-499 3A=4100-1 => A=(4100-1)/3 Vì 4100>4100-1 nên (4100-1)/3 < 4100/3 HAY A<B/3(ĐPCM)
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Bài làm:
Ta có: \(A=1+4+4^2+4^3+...+4^{99}\)
\(\Rightarrow4A=4+4^2+4^3+4^4+...+4^{100}\)
\(\Rightarrow4A-A=\left(4+4^2+...+4^{100}\right)-\left(1+4+...+4^{99}\right)\)
\(\Leftrightarrow3A=4^{100}-1\)
\(\Rightarrow A=\frac{4^{100}-1}{3}=\frac{4^{100}}{3}-\frac{1}{3}< \frac{4^{100}}{3}=\frac{B}{3}\)
\(\Leftrightarrow A< \frac{B}{3}\)
A = 1 + 4 + 42 + 43 + ... + 499
4A = 4( 1 + 4 + 42 + 43 + ... + 499 )
= 4 + 42 + 43 + 44 + ... + 4100
4A - A = 3A
= ( 4 + 42 + 43 + 44 + ... + 4100 ) - ( 1 + 4 + 42 + 43 + ... + 499 )
= 4 + 42 + 43 + 44 + ... + 4100 - 1 - 4 - 42 - 43 - ... - 499
= 4100 - 1
3A = 4100 - 1 => A = \(\frac{4^{100}-1}{3}\)
\(\frac{B}{3}=\frac{4^{100}}{3}\)
\(4^{100}-1< 4^{100}\Rightarrow\frac{4^{100}-1}{3}< \frac{4^{100}}{3}\)
\(\Rightarrow A< \frac{B}{3}\left(đpcm\right)\)
\(A=1+4+4^2+...+4^{99}\)
=>\(4A=4+4^2+4^3+...+4^{100}\)
=>\(4A-A=\left(4+4^2+4^3+...+4^{100}\right)-\left(1+4+4^2+...+4^{98}\right)\)
=>\(3A=4^{100}-1\)
=>\(A=\frac{4^{100}-1}{3}< \frac{4^{100}}{3}=\frac{B}{3}\)
Ta có đpcm
4A=4+42+43+44+...+499+4100
=> 4A-A=4+42+43+44+...+499+4100-(1+4+42+43+44+...+499)=4100-1
=> 3A=4100-1 => A=\(\frac{4^{100}-1}{3}=\frac{4^{100}}{3}-\frac{1}{3}=\frac{B}{3}-\frac{1}{3}\)
=> A < B/3