7. 求$1 + 2 + 2^{2}+2^{3}+…+2^{2016}$的值,可令$S = 1 + 2 + 2^{2}+2^{3}+…+2^{2016}$,则$2S = 2 + 2^{2}+2^{3}+…+2^{2017}$,因此$2S - S = 2^{2017}-1$,$S = 2^{2017}-1$.参照以上过程,计算$S = 3 + 3^{2}+3^{3}+…+3^{2022}+3^{2023}$的值为
$\frac{3^{2024}-3}{2}$
.答案
$\frac{3^{2024}-3}{2}$
解析
令$S = 3 + 3^{2} + 3^{3} + \ldots + 3^{2022} + 3^{2023}$,
则$3S = 3^{2} + 3^{3} + \ldots + 3^{2023} + 3^{2024}$,
$3S - S = (3^{2} + 3^{3} + \ldots + 3^{2023} + 3^{2024}) - (3 + 3^{2} + 3^{3} + \ldots + 3^{2022} + 3^{2023})$,
$2S = 3^{2024} - 3$,
$S = \frac{3^{2024} - 3}{2}$。
则$3S = 3^{2} + 3^{3} + \ldots + 3^{2023} + 3^{2024}$,
$3S - S = (3^{2} + 3^{3} + \ldots + 3^{2023} + 3^{2024}) - (3 + 3^{2} + 3^{3} + \ldots + 3^{2022} + 3^{2023})$,
$2S = 3^{2024} - 3$,
$S = \frac{3^{2024} - 3}{2}$。
8. 计算:
(1)$\frac{4}{9}-\frac{1}{4}×(-\frac{2}{3})^{2}+\frac{1}{6}÷(-1.5)^{2}$;
(2)$(-\frac{1}{3}+2)^{2}-4×(-\frac{1}{3})^{2}÷\frac{(-2)^{2}}{3}$;
(3)$(-1)^{3}-(1 - 0.5)÷\{\frac{1}{3}×[2 - (-3)^{2}]\}$;
(4)$-1^{2024}+(-2)^{3}×(-\frac{1}{4})+(-3)^{2}-|-1 - 3|$.
(1)$\frac{4}{9}-\frac{1}{4}×(-\frac{2}{3})^{2}+\frac{1}{6}÷(-1.5)^{2}$;
(2)$(-\frac{1}{3}+2)^{2}-4×(-\frac{1}{3})^{2}÷\frac{(-2)^{2}}{3}$;
(3)$(-1)^{3}-(1 - 0.5)÷\{\frac{1}{3}×[2 - (-3)^{2}]\}$;
(4)$-1^{2024}+(-2)^{3}×(-\frac{1}{4})+(-3)^{2}-|-1 - 3|$.
答案
(1) $\frac{11}{27}$
(2) $\frac{22}{9}$
(3) $-\frac{11}{14}$
(4) 6
9. 观察下面三行数:
①$2$,$-4$,$8$,$-16$,$32$,$-64$,…;
②$4$,$-2$,$10$,$-14$,$34$,$-62$,…;
③$-1$,$2$,$-4$,$8$,$-16$,$32$,…;
取每一行的第$n$个数,依次记为$a$,$b$,$c$.当$n = 2$时,$a = -4$,$b = -2$,$c = 2$.
(1)当$n = 8$时,$a = $
(2)已知$n$为奇数,且$a$,$b$,$c$这三个数中,最大的数与最小的数的差为$194$,求$n$的值.
①$2$,$-4$,$8$,$-16$,$32$,$-64$,…;
②$4$,$-2$,$10$,$-14$,$34$,$-62$,…;
③$-1$,$2$,$-4$,$8$,$-16$,$32$,…;
取每一行的第$n$个数,依次记为$a$,$b$,$c$.当$n = 2$时,$a = -4$,$b = -2$,$c = 2$.
(1)当$n = 8$时,$a = $
-256
,$b = $-254
,$c = $128
,这三个数中最大的数与最小的数的差为384
;(2)已知$n$为奇数,且$a$,$b$,$c$这三个数中,最大的数与最小的数的差为$194$,求$n$的值.
7
答案
(1) -256 -254 128 384
(2) 7
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