已知数列{an},a3=3,a7=5,且an+2+an=2an+1,求a1的值 (n,n+1,n+2都是下标符号)
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已知数列{an},a3=3,a7=5,且an+2+an=2an+1,求a1的值 (n,n+1,n+2都是下标符号)
我已经想的不行了! 求破啊!
an+2+an=2an+1
移项
an+2 - an+1 = an+1 -an
说明{ an+1 -an} 为等差数列.设公差为d
a7 - a3 = (a7-a6) +(a6-a5) + (a5-a4) +(a4-a3) = 4d
d = (5-3)/4 = 1/2
a3-a1 = (a3-a2) + (a2-a1) = 2d = 1
a1 = a3-1 = 3-1 =2
移项
an+2 - an+1 = an+1 -an
说明{ an+1 -an} 为等差数列.设公差为d
a7 - a3 = (a7-a6) +(a6-a5) + (a5-a4) +(a4-a3) = 4d
d = (5-3)/4 = 1/2
a3-a1 = (a3-a2) + (a2-a1) = 2d = 1
a1 = a3-1 = 3-1 =2
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