11. (42分)计算：
(1)$13+59.8-12\frac{1}{5}-30\frac{4}{5}-8.1$ (2)$-2^4-(-3+7)^2-(-1)^2×(-2)$
(3)$-81÷2\frac{1}{4}×|-\frac{4}{9}|-28÷(-\frac{7}{4})$ (4)$(-2×\frac{1}{2})^{101}×3+[(-3)^3-3^3]$
(5)$(-10)÷(-\frac{1}{2})-(-10)×\frac{1}{2}+2×(-10)$
(6)$-1^3-(-\frac{1}{4}+\frac{5}{6}+\frac{8}{9})÷\frac{1}{36}×(-4)$

（1）21.7 （2）-30 （3）0 （4）-57 （5）5 （6）211

(1) $13 + 59.8 - 12\frac{1}{5} - 30\frac{4}{5} - 8.1$
$=13 + 59.8 - 12.2 - 30.8 - 8.1$
$=(13) + (59.8 - 12.2 - 30.8) - 8.1$
$=13 + 16.8 - 8.1$
$=29.8 - 8.1$
$=21.7$
(2) $-2^4 - (-3 + 7)^2 - (-1)^2×(-2)$
$=-16 - 4^2 - 1×(-2)$
$=-16 - 16 + 2$
$=-32 + 2$
$=-30$
(3) $-81÷2\frac{1}{4}×|-\frac{4}{9}| - 28÷(-\frac{7}{4})$
$=-81÷\frac{9}{4}×\frac{4}{9} - 28×(-\frac{4}{7})$
$=-81×\frac{4}{9}×\frac{4}{9} + 16$
$=-36×\frac{4}{9} + 16$
$=-16 + 16$
$=0$
(4) $(-2×\frac{1}{2})^{101}×3 + [(-3)^3 - 3^3]$
$=(-1)^{101}×3 + (-27 - 27)$
$=-1×3 + (-54)$
$=-3 - 54$
$=-57$
(5) $(-10)÷(-\frac{1}{2}) - (-10)×\frac{1}{2} + 2×(-10)$
$=(-10)×(-2) + 5 - 20$
$=20 + 5 - 20$
$=25 - 20$
$=5$
(6) $-1^3 - (-\frac{1}{4} + \frac{5}{6} + \frac{8}{9})÷\frac{1}{36}×(-4)$
$=-1 - (-\frac{1}{4} + \frac{5}{6} + \frac{8}{9})×36×(-4)$
$=-1 - [(-\frac{1}{4})×36 + \frac{5}{6}×36 + \frac{8}{9}×36]×(-4)$
$=-1 - (-9 + 30 + 32)×(-4)$
$=-1 - 53×(-4)$
$=-1 + 212$
$=211$

12. (18分)为了求$1+3+3^2+3^3+…+3^{100}$的值,可设$M= 1+3+3^2+3^3+…+3^{100}$,则$3M= 3+3^2+3^3+3^4+…+3^{101}$,两式相减,得$2M= 3^{101}-1$,所以$M= \frac{3^{101}-1}{2}$,即$1+3+3^2+3^3+…+3^{100}= \frac{3^{101}-1}{2}$.仿照以上过程,计算：$1+5+5^2+5^3+…+5^{205}$.

设$N=1+5+5^{2}+5^{3}+\cdots+5^{205}$,则$5N=5+5^{2}+5^{3}+5^{4}+\cdots+5^{206}$,两式相减,得$4N=5^{206}-1$,所以$N=\frac{5^{206}-1}{4}$,即$1+5+5^{2}+5^{3}+\cdots+5^{205}=\frac{5^{206}-1}{4}$