已知等差数列公差为d,1/a1a2+1/a2a3+…+1/anan+1可化简为已知等差数列公差为d,且a1≠0,d≠0,则1/a1a2+1/a2a3+…+1/anan+1可化简为 n/a1(a1+nd)求详细步骤
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![已知等差数列公差为d,1/a1a2+1/a2a3+…+1/anan+1可化简为已知等差数列公差为d,且a1≠0,d≠0,则1/a1a2+1/a2a3+…+1/anan+1可化简为 n/a1(a1+nd)求详细步骤](/uploads/image/z/1751585-41-5.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%E5%85%AC%E5%B7%AE%E4%B8%BAd%2C1%2Fa1a2%2B1%2Fa2a3%2B%E2%80%A6%2B1%2Fanan%2B1%E5%8F%AF%E5%8C%96%E7%AE%80%E4%B8%BA%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%E5%85%AC%E5%B7%AE%E4%B8%BAd%2C%E4%B8%94a1%E2%89%A00%2Cd%E2%89%A00%2C%E5%88%991%2Fa1a2%2B1%2Fa2a3%2B%E2%80%A6%2B1%2Fanan%2B1%E5%8F%AF%E5%8C%96%E7%AE%80%E4%B8%BA+n%2Fa1%28a1%2Bnd%29%E6%B1%82%E8%AF%A6%E7%BB%86%E6%AD%A5%E9%AA%A4)
已知等差数列公差为d,1/a1a2+1/a2a3+…+1/anan+1可化简为已知等差数列公差为d,且a1≠0,d≠0,则1/a1a2+1/a2a3+…+1/anan+1可化简为 n/a1(a1+nd)求详细步骤
已知等差数列公差为d,1/a1a2+1/a2a3+…+1/anan+1可化简为
已知等差数列公差为d,且a1≠0,d≠0,则1/a1a2+1/a2a3+…+1/anan+1可化简为 n/a1(a1+nd)求详细步骤
已知等差数列公差为d,1/a1a2+1/a2a3+…+1/anan+1可化简为已知等差数列公差为d,且a1≠0,d≠0,则1/a1a2+1/a2a3+…+1/anan+1可化简为 n/a1(a1+nd)求详细步骤
因为1/anan+1=1/an*(an+d)=1/d[1/an-1/(an+d)]=1/d[1/an-1/an+1]
所以1/a1a2+1/a2a3+…+1/anan+1
=1/d[1/a1-1/a2+1/a3-1/a4.1/an-1/an+1]
=1/d(1/a1-1/an+1)
=1/d*(an+1-a1)/a1an+1
=n/a1(a1+nd)