已知数列an的奇数项是公差d1的等差数列,偶数项是公差为d2的等差数列 Sn是前n项和,a1=1,a2=2已知S15=15a8,且对任意n,有an这个不要复制了,是错的!***************************************************************
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![已知数列an的奇数项是公差d1的等差数列,偶数项是公差为d2的等差数列 Sn是前n项和,a1=1,a2=2已知S15=15a8,且对任意n,有an这个不要复制了,是错的!***************************************************************](/uploads/image/z/5911740-36-0.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%E7%9A%84%E5%A5%87%E6%95%B0%E9%A1%B9%E6%98%AF%E5%85%AC%E5%B7%AEd1%E7%9A%84%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%81%B6%E6%95%B0%E9%A1%B9%E6%98%AF%E5%85%AC%E5%B7%AE%E4%B8%BAd2%E7%9A%84%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97+Sn%E6%98%AF%E5%89%8Dn%E9%A1%B9%E5%92%8C%2Ca1%3D1%2Ca2%3D2%E5%B7%B2%E7%9F%A5S15%3D15a8%2C%E4%B8%94%E5%AF%B9%E4%BB%BB%E6%84%8Fn%2C%E6%9C%89an%E8%BF%99%E4%B8%AA%E4%B8%8D%E8%A6%81%E5%A4%8D%E5%88%B6%E4%BA%86%EF%BC%8C%E6%98%AF%E9%94%99%E7%9A%84%EF%BC%81%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A%2A)
已知数列an的奇数项是公差d1的等差数列,偶数项是公差为d2的等差数列 Sn是前n项和,a1=1,a2=2已知S15=15a8,且对任意n,有an这个不要复制了,是错的!***************************************************************
已知数列an的奇数项是公差d1的等差数列,偶数项是公差为d2的等差数列 Sn是前n项和,a1=1,a2=2
已知S15=15a8,且对任意n,有an
这个不要复制了,是错的!
***********************************************************************************************************
2、根据题意,有
S15=15a8=15*(a8+a8)/2,假若a7=a8-d,a9=a8+d,那么S15=15*(a1+a15)/2,这个就是等差数列求和所以,数列an为等差数列
已知数列an的奇数项是公差d1的等差数列,偶数项是公差为d2的等差数列 Sn是前n项和,a1=1,a2=2已知S15=15a8,且对任意n,有an这个不要复制了,是错的!***************************************************************
and1=d2
若d1>d2
a(2n-1)-a(2n)
=1+(2n-2)/2*d1-2-(2n-2)/2*d2
=(2n-2)/2*(d1-d2)-1
当n不断增大则a(2n-1)-a(2n)>0 不成立
若d10 不成立
∴d1=d2
又∵S15=15a8
S15=(a1+a3+...+a15)+(a2+a4+...+a14)
=8a1+32d1+7a2+21d2
=22+32d1+21d2=15a8=15(a2+3d2)=30+45d2
==>32d1=24d2+8
又∵d1=d2 ∴d1=d2=1
∴数列an是公差为1的等差数列
已知数列an的奇数项是公差d1的等差数列,偶数项是公差为d2的等差数列 Sn是前n项和,a1=1,a2=2
1.若n为偶数,则:
Sn=a1+a3+...+a(n-1)+a2+a4+...+an
=n/2*a1+n/2*(n/2-1)*d1/2+n/2*a2+n/2*(n/2-1)*d2/2
=3/2n+n(n-2)(d1+d2)/8
2.若...
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已知数列an的奇数项是公差d1的等差数列,偶数项是公差为d2的等差数列 Sn是前n项和,a1=1,a2=2
1.若n为偶数,则:
Sn=a1+a3+...+a(n-1)+a2+a4+...+an
=n/2*a1+n/2*(n/2-1)*d1/2+n/2*a2+n/2*(n/2-1)*d2/2
=3/2n+n(n-2)(d1+d2)/8
2.若n为奇数,则:
Sn=a1+a3+...+an+a2+a4+...+a(n-1)
=(n+1)/2*a1+(n+1)/2*[(n+1)/2-1]*d1/2+(n-1)/2*a2+(n-1)/2*[(n-1)/2-1]*d2/2
=3/2n-1/2+1/8(n-1)(n+1)d1+1/8(n-1)(n-3)d2
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