f.3指數定律

1. [3^(n+1)-3^(n)]
=3^n * 3^1 – 3^n
=3
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2. [3^(n+1)-3^(n-1)]
=3^n * 3^1 - (3^n * 3^-1)
=3^n * 3^1 – 3^n * -1/3
=3 * -1/3
=-1
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3. 5^(n+2)-5n
=5^n + 5^2 – 5n
=25 + 5^n -5n
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4. 試按數值遞升次序排列: 3^(729), 9^(237), 27^(39), 729^(3)
3^(729)
9^(237)=[(3^2)^237]=3^(2*237)=3^474
27^(39)= [(3^3)^39] =3^(3*39)=3^117
729^(3)= [(3^6)^3] =3^(6*3)=3^18
所以729^(3) , 27^(39) , 9^(237) , 3^(729)

p.s. 指數愈大, 數值愈大。

1. 3^(n+1)-3^(n) = 3[3^(n)] -3^(n) = 2[3^(n)]

2. 3^(n+1)-3^(n-1) =3[ 3(3^(n-1)] -3^(n-1) =8[3^(n-1)]

3. 5^(n+2)-5n = 5[5^(n+1)]- 5n = 5[5^(n+1)- n]

4. 3^(729)
9^(237) =3^2^(237)= 3^[2(237)]= 3^(474)
27^(39) =3^3^(39) =3^[3(39)] =3^(117)
729^(3) =(27^2)^(3) =27^[2(3)] =27^(6) =(3^3)^(6)= 3^[3(6)]=3^(18)

729^(3) < 27^(39) < 9^(237) < 3^(729)