数列an,a1=1,n>=2,an=(根号下sn +根号sn-1)/2,求数列根号sn为等差数列,及an通项我想问的是,“(2)由(1)得√Sn=1+(n-1)/2=(n+1)/2”这里是如何得出的?
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![数列an,a1=1,n>=2,an=(根号下sn +根号sn-1)/2,求数列根号sn为等差数列,及an通项我想问的是,“(2)由(1)得√Sn=1+(n-1)/2=(n+1)/2”这里是如何得出的?](/uploads/image/z/3927924-36-4.jpg?t=%E6%95%B0%E5%88%97an%2Ca1%3D1%2Cn%3E%3D2%2Can%3D%28%E6%A0%B9%E5%8F%B7%E4%B8%8Bsn+%2B%E6%A0%B9%E5%8F%B7sn-1%29%2F2%2C%E6%B1%82%E6%95%B0%E5%88%97%E6%A0%B9%E5%8F%B7sn%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%8F%8Aan%E9%80%9A%E9%A1%B9%E6%88%91%E6%83%B3%E9%97%AE%E7%9A%84%E6%98%AF%2C%E2%80%9C%EF%BC%882%EF%BC%89%E7%94%B1%EF%BC%881%EF%BC%89%E5%BE%97%E2%88%9ASn%3D1%2B%EF%BC%88n-1%29%2F2%3D%28n%2B1%29%2F2%E2%80%9D%E8%BF%99%E9%87%8C%E6%98%AF%E5%A6%82%E4%BD%95%E5%BE%97%E5%87%BA%E7%9A%84%3F)
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数列an,a1=1,n>=2,an=(根号下sn +根号sn-1)/2,求数列根号sn为等差数列,及an通项我想问的是,“(2)由(1)得√Sn=1+(n-1)/2=(n+1)/2”这里是如何得出的?
数列an,a1=1,n>=2,an=(根号下sn +根号sn-1)/2,求数列根号sn为等差数列,及an通项
我想问的是,“(2)由(1)得√Sn=1+(n-1)/2=(n+1)/2”这里是如何得出的?
数列an,a1=1,n>=2,an=(根号下sn +根号sn-1)/2,求数列根号sn为等差数列,及an通项我想问的是,“(2)由(1)得√Sn=1+(n-1)/2=(n+1)/2”这里是如何得出的?
数列an,a1=1,n>=2,an=(根号下sn +根号sn-1)/2(下面√表示根号)
Sn-Sn-1=an=1/2(√Sn+√Sn-1)
(√Sn-√Sn-1-1/2)(√Sn+√Sn-1)=0
√Sn-√Sn-1=1/2
所以√Sn是以√1为首项1/2为公差的等差数列
√Sn=1+1/2(n-1)=1/2(n+1) (等差数列的通项公式an=a1+(n-1)*d)
n>=2
an=1/2(√Sn+√Sn-1)=1/2*n+1/4
所以 1 n=1
an=
1/2*n+1/4 n>=2
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