已知等差数列公差为d,1/a1a2+1/a2a3+…+1/anan+1可化简为
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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
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)
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)
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