化简 ,下面的数都在二次根号内,(n为正整数),1/n^2+1/(n+1)^2+1/(2n+1)^2,
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![化简 ,下面的数都在二次根号内,(n为正整数),1/n^2+1/(n+1)^2+1/(2n+1)^2,](/uploads/image/z/7617296-56-6.jpg?t=%E5%8C%96%E7%AE%80+%2C%E4%B8%8B%E9%9D%A2%E7%9A%84%E6%95%B0%E9%83%BD%E5%9C%A8%E4%BA%8C%E6%AC%A1%E6%A0%B9%E5%8F%B7%E5%86%85%2C%28n%E4%B8%BA%E6%AD%A3%E6%95%B4%E6%95%B0%29%2C1%2Fn%5E2%2B1%2F%28n%2B1%29%5E2%2B1%2F%282n%2B1%29%5E2%2C)
化简 ,下面的数都在二次根号内,(n为正整数),1/n^2+1/(n+1)^2+1/(2n+1)^2,
化简 ,下面的数都在二次根号内,(n为正整数),1/n^2+1/(n+1)^2+1/(2n+1)^2,
化简 ,下面的数都在二次根号内,(n为正整数),1/n^2+1/(n+1)^2+1/(2n+1)^2,
设a=n,b=n+1,有b-a=1
则通分得分子为:
(ab)^2+a^2(a+b)^2+b^2(a+b)^2
=a^2b^2+(a+b)^2(a^2+b^2)
=a^2b^2+[(a-b)^2+4ab][(a-b)^2+2ab]
=a^2b^2+(1+4ab)(1+2ab)
=a^2b^2+1+6ab+8a^2b^2
=9a^2b^2+6ab+1
=(3ab+1)^2
所以原根式=(3ab+1)/[ab(a+b)]=[3n(n+1)+1]/[n(n+1)/(2n+1)]