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**********TÍNH HỢP LÍ
Đặt a=1+3+3^2+....+3^99
Suy ra 3a=3*(1+3+3^2+.......+3^9)
Suy ra 3a=3+3^2+3^3+.......+3^100
Suy ra 3a-a=(3+3^2+3^3+.......+3^100)-(1+3+3^2+3^3+.......3^99)
Suy ra 2a=3^100-1
Suy ra a=(3^100-1)/2
Suy ra (1+3+3^2+......+3^99)*2
=3^100-1
Suy ra 3^100-1+1
= 3^100
Suy ra (1+3+3^2+.......+3^99)*2+1
=3^100
1.(5.3^11+4.3^12):(3^9.5^2-3^9.2^3)
=(5.3^11+4.3^12):(3^9.5^2-3^9.2^3)
=(5.3^11+4.3^11.3):[3^9.(5^2-2^3)]
=(5.3^11+12.3^11):[3^9.17]
=3^11.(5+12):(3^9.17)
=(3^11.17):(3^9.17)
=17.(3^11:3^9)
=17.3^2
=17.9
=153
2.(12+22+32+...+992+1002).(36.333-108.111)
=2.(12+22+32+...+992+1002).(36.333-36.3.111)
=2.(12+22+32+...+992+1002).(36.333-36.333)
=2.(12+22+32+...+992+1002).0
=0
1) 3F=3+32+33+34+...+32016
3F-F=(3+32+33+34+...+32016)-( 1+3+32+33+...+32015)
2F=32016-1
F= 32016-1/2...
2)
D=3+33+35+37+.....+399
=>32D=33+35+37+39+.....+3101
=>32D-D=33+35+37+39+.....+3101-3-33-35-37-....-399
=>8D=3101-3
=>D=(3101-3):8
Chơi câu khó nhất
D = 4 + 42 + 43 + ... + 4n
4D = 42 + 43 + ... + 4n+1
3D = 4n+1 - 4
D = \(\frac{4^{n+1}-4}{3}\)
Đặt A = 1 + 2 + 22 + 23 + 24 + ... + 299 + 2100
=> 2A = 2( 1 + 2 + 22 + 23 + 24 + ... + 299 + 2100 )
= 2 + 22 + 23 + 24 + ... + 2100 + 2101
=> A = 2A - A
= 2 + 22 + 23 + 24 + ... + 2100 + 2101 - ( 1 + 2 + 22 + 23 + 24 + ... + 299 + 2100 )
= 2 + 22 + 23 + 24 + ... + 2100 + 2101 - 1 - 2 - 22 - 23 - 24 - ... - 299 - 2100
= 2101 - 1
Đătj S= 1+2+22+23+24+.......+299+2100
\(\Rightarrow2S=2+2^2+2^3+...+2^{101}\)
\(\Rightarrow2S-S=(2+2^2+2^3+...+2^{101})-\)\((1+2+2^2+...+2^{100})\)
\(\Rightarrow S=2^{101}-1\)
\(a,\)Đặt \(A=1+2+2^2+...+2^{99}+2^{100}\)
\(\Rightarrow2A=2+2^2+...+2^{100}+2^{101}\)
\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{101}\right)-\left(1+2+2^2+...2^{100}\right)\)
\(\Rightarrow A=2^{101}-1\)
\(b,\)Đặt \(B=5+5^3+5^5+...+5^{97}+5^{99}\)
\(\Rightarrow5^2B=5^3+5^5+...+5^{99}+5^{101}\)
\(\Rightarrow25B-B=\left(5^3+5^5+...+5^{99}+5^{101}\right)-\left(5+5^3+...+5^{99}\right)\)
\(\Rightarrow24B=5^{101}-5\)
\(\Rightarrow B=\frac{5^{101}-5}{24}\)
Đặt A=(1+3+32+33+...+399) .2+1
Đặt B=1+3+32+33+...+399
3B=3+32+33+34+...+3100
3B-B=(3+32+33+34+...+3100)-(1+3+32+33+...+399)
2B = 3100 - 1
B = \(\frac{3^{100}-1}{2}\)
Thay B vào A ta được:
\(A=\frac{3^{100}-1}{2}.2+1=3^{100}-1+1=3^{100}\)