16. 春秋时期,孔子有一天对他的弟子们说道:“不愤不启,不悱不发,举一隅不以三隅反,则不复也.”这句话至今仍被视为教育学的黄金法则. 例如,在数学教学中,老师常通过设置问题链引导学生逐步推导公式,而非直接告知结论,正是对这一理念的实践. 请阅读下面的解题过程,感受从特殊到一般的数学思想,类比推理解决以下问题.
例题:化简$(2+1)(2^2+1)(2^4+1)$.
解:原式$=(2-1)(2+1)(2^2+1)(2^4+1)$
$=(2^2-1)(2^2+1)(2^4+1)$
$=(2^4-1)(2^4+1)$
$=2^8 - 1$.
(1)填空:$\_\_\_\_\_\_ × (b+a)=a^2 - b^2$;
(2)化简:$(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$;
(3)运用上面所学内容化简下列式子:
①$(5-1)(5^2+1)(5^4+1)···(5^{128}+1)$;
②$(m+1)(m^2+1)(m^4+1)···(m^{2^n}+1)$,其中$m(m≠1)$,$n$均为正整数.
例题:化简$(2+1)(2^2+1)(2^4+1)$.
解:原式$=(2-1)(2+1)(2^2+1)(2^4+1)$
$=(2^2-1)(2^2+1)(2^4+1)$
$=(2^4-1)(2^4+1)$
$=2^8 - 1$.
(1)填空:$\_\_\_\_\_\_ × (b+a)=a^2 - b^2$;
(2)化简:$(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$;
(3)运用上面所学内容化简下列式子:
①$(5-1)(5^2+1)(5^4+1)···(5^{128}+1)$;
②$(m+1)(m^2+1)(m^4+1)···(m^{2^n}+1)$,其中$m(m≠1)$,$n$均为正整数.
答案
16. (1)$(a - b) × (b + a) = a^2 - b^2$,故答案为$(a - b).$
(2)原式$=\frac{1}{2}(3 - 1)(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^{16} + 1)$
$=\frac{1}{2}(3^2 - 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^{16} + 1)$
$=\frac{1}{2}(3^4 - 1)(3^4 + 1)(3^8 + 1)(3^{16} + 1)$
$=\frac{1}{2}(3^8 - 1)(3^8 + 1)(3^{16} + 1)$
$=\frac{1}{2}(3^{16} - 1)(3^{16} + 1)$
$=\frac{1}{2}(3^{32} - 1).$
(3)①原式$=\frac{1}{6}(5 + 1)(5 - 1)(5^2 + 1)(5^4 + 1)···(5^{128} + 1)$
$=\frac{1}{6}(5^2 - 1)(5^2 + 1)(5^4 + 1)···(5^{128} + 1)$
$=\frac{1}{6}(5^4 - 1)(5^4 + 1)···(5^{128} + 1)$
$=\frac{1}{6}(5^8 - 1)···(5^{128} + 1)$
$=\frac{1}{6}(5^{256} - 1),$
②原式$=\frac{1}{m - 1}(m - 1)(m + 1)(m^2 + 1)(m^4 + 1)···(m^{2^n} + 1)$
$=\frac{1}{m - 1}(m^2 - 1)(m^2 + 1)(m^4 + 1)···(m^{2^n} + 1)$
$=\frac{1}{m - 1}(m^4 - 1)(m^4 + 1)···(m^{2^n} + 1)$
$=\frac{1}{m - 1}(m^8 - 1)···(m^{2^n} + 1)$
$=\frac{1}{m - 1}(m^{2^{n + 1}} - 1),$
(2)原式$=\frac{1}{2}(3 - 1)(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^{16} + 1)$
$=\frac{1}{2}(3^2 - 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^{16} + 1)$
$=\frac{1}{2}(3^4 - 1)(3^4 + 1)(3^8 + 1)(3^{16} + 1)$
$=\frac{1}{2}(3^8 - 1)(3^8 + 1)(3^{16} + 1)$
$=\frac{1}{2}(3^{16} - 1)(3^{16} + 1)$
$=\frac{1}{2}(3^{32} - 1).$
(3)①原式$=\frac{1}{6}(5 + 1)(5 - 1)(5^2 + 1)(5^4 + 1)···(5^{128} + 1)$
$=\frac{1}{6}(5^2 - 1)(5^2 + 1)(5^4 + 1)···(5^{128} + 1)$
$=\frac{1}{6}(5^4 - 1)(5^4 + 1)···(5^{128} + 1)$
$=\frac{1}{6}(5^8 - 1)···(5^{128} + 1)$
$=\frac{1}{6}(5^{256} - 1),$
②原式$=\frac{1}{m - 1}(m - 1)(m + 1)(m^2 + 1)(m^4 + 1)···(m^{2^n} + 1)$
$=\frac{1}{m - 1}(m^2 - 1)(m^2 + 1)(m^4 + 1)···(m^{2^n} + 1)$
$=\frac{1}{m - 1}(m^4 - 1)(m^4 + 1)···(m^{2^n} + 1)$
$=\frac{1}{m - 1}(m^8 - 1)···(m^{2^n} + 1)$
$=\frac{1}{m - 1}(m^{2^{n + 1}} - 1),$
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