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2001²-2000²+1999²-1998²+…+3²-2²+1²=(写出过程,递等式计算)
人气:488 ℃ 时间:2019-11-24 04:14:50
解答
使用 (a + b)(a - b) = a² - b²
2001² - 2000² + 1999² - 1998² + ...+ 3² - 2² + 1²
= (2001² - 2000²) + (1999² - 1998²) + ...+ (3² - 2²) + 1²
= (2001 + 2000)(2001 - 2000) + (1999 + 1998)(1999 - 1998) + ...+ (3 + 2)(3 - 2) + 1
= (2001 + 2000) * 1 + (1999 + 1998) * 1 + ...+ (3 + 2) * 1 + 1
= 2001 + 2000 + 1999 + 1998 + .+ 3 + 2 + 1
= (1 + 2001) * 2001 / 2 (梯形公式)
= 2003001为什么(a+b)(a-b)=a²-b²?公式呀!
(a + b) (a - b) = a(a + b) - b(a + b) = a² + ab - ab - b² = a² - b²
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