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Lời giải:
$T=3-3^2+3^3-3^4+....-3^{2000}$
$3T=3^2-3^3+3^4-3^5+...-3^{2001}$
$\Rightarrow T+3T=3-3^{2001}$
$\Rightarrow 4T=3-3^{2001}$
$\Rightarrow T=\frac{3-3^{2001}}{4}$
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B = 3 - 32 + 33 - ..... + 31999 - 32000
=> 3B = 32 - 33 + 34 - ..... + 32000 - 32001
=> 3B + B = 4B = 3 - 32001
=> 32001 = 3 - 4B
=> n = 2001
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chac o trong cau hoi tuong tu co day
3B=32-33+34-35+..............+32000-32001
3B+B=(32-33+34-35+............+32000-32001)-(3-32+33-34+............+31999-32000)
3B+B=3-32001
=>4B=3-32001
=>3-4B=-32001
=>n=2001
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Ta có:
B = 3 - 32 + 33 - 34 + ...... + 31999 - 32000
=> 3B = 32 - 33 + 34 - 35 + ...... + 32000 - 32001
=> 3B + B = 4B = 3 - 32001
=> 32001 = 3 - 4B
Vậy n = 2001