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Giải : S = 2.(2 + 0 ) + 4.(3 + 1) + 6.(4 + 2) + .... + 100.(
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Đặt A= 1/22+1/24+...+1/2100
Ta có 4A=1+1/22+1/24+..+1/298
=> 4A-A=(1+1/22+1/24+..+1/298)-(1/22+1/24+...+1/2100)
=> 3A=1-1/2100 =>A=\(\frac{1-\frac{1}{2^{100}}}{3}\)
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S = 1 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + ... + 2^100
2S = 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + ... + 2^100 + 2^101
2S - S = ( 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + ... + 2^100 + 2^101 ) - ( 1 + 2 + 2^2 +2^3 +2^4 + 2^5 + .... + 2^100 )
S = 2^101 - 1
Vậy S = 2^101 - 1
Ta có :
S = 1 + 2 + 22 + 23 + 24 + 25 + ... + 2100
2S = 2 + 22 + 23 + 24 + 25 + ... + 2101
2S - S = ( 2 + 22 + 23 + 24 + 25 + ... + 2101 ) - ( 1 - 2 - 22 - 23 - 24 - 25 - ... - 2100 )
S = 2101 - 1
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\(S=100^2-99^2+...+2^2-1^2=\left(100+99\right)+\left(98+97\right)+..+\left(2+1\right)\)
\(S=100+99+..+2+1\)
\(S=1+2+..+99+100\)
\(2S=\left(1+100\right)+..+\left(1+100\right)\)
\(S=\frac{100.\left(100+1\right)}{2}=50.101\)
S=(22+42+62+.......+1002)-(12+32+52+......+992)
S=22+42+62+.....+1002-12+32+52+.....+992
S=(22-12)+(42-32)+.........+(1002-992)
Sử dụng công thức a2-b2=(a+b)(a-b)
S=(2+1)(2-1)+(4+3)(4-3)+.......+(100+99)(100-99)
S=3.1+7.1+.......+199.1
s=3+7+........+199
tính S =5050
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S=1+22+24+...+2100
4S=22B=22+24+26+...+2102
3B=4B-B=2102-1
=> B = \(\frac{2^{102}-1}{3}\)
\(S=1+2^2+2^4+2^6+...+2^{100}\\ 2^2S=2^2+2^4+2^6+2^8+....+2^{102}\\ 2S-S=S=\left(2^2+2^4+2^6+2^8+....+2^{102}\right)-\left(1+2^2+2^4+2^6+...+2^{100}\right)\\ S=2^{102}-1\)