Question

 閱讀下列材料:

某同學(xué)在計(jì)算3×(4+1)(4²+1)時(shí),把3寫成4-1后,發(fā)現(xiàn)可以連續(xù)運(yùn)用平方差公式計(jì)算:3×(4+1)(4²+1)=(4-1)(4+1)(4²+1)=(4²-1)(4²+1)=16²-1.很受啟發(fā),后來(lái)在求(2+1)(2²+1)(2⁴+1)(2⁸+1)…(2¹⁰²⁴+1)的值時(shí),又改造此法,將乘積式前面乘以1,且把1寫為2-1得(2+1)(2²+1)(2⁴+1)(2⁸+1)…(2¹⁰²⁴+1)

=(2-1)(2+1)(2²+1)(2⁴+1)(2⁸+1)…(2¹⁰²⁴+1)

=(2²-1)(2²+1)(2⁴+1)(2⁸+1)…(2¹⁰²⁴+1)

=(2⁴-1)(2⁴+1)(2⁸+1)…(2¹⁰²⁴+1)=…

=(2¹⁰²⁴-1)(2¹⁰²⁴+1)=2²⁰⁴⁸-1.

回答下列問題:

(1)請(qǐng)借鑒該同學(xué)的經(jīng)驗(yàn),計(jì)算:

(3+1)(3²+1)(3⁴+1)(3⁸+1).

(2)借用上面的方法,再逆用平方差公式計(jì)算:

….

Answer 1

【解析】(1)(3+1)(3²+1)(3⁴+1)(3⁸+1)

=(3²-1)(3²+1)(3⁴+1)(3⁸+1)

=(3⁴-1)(3⁴+1)(3⁸+1)=(3⁸-1)(3⁸+1)

=(3¹⁶-1).

(2)…

=…

=××××…××

=×=.