求数列前N项和Tn,An = n -1,n属于正整数 求Tn = (a1)^2 -(a2)^2+(a3)^2-(a4)^2.+(-1)^(n+1) * bn^2
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![求数列前N项和Tn,An = n -1,n属于正整数 求Tn = (a1)^2 -(a2)^2+(a3)^2-(a4)^2.+(-1)^(n+1) * bn^2](/uploads/image/z/5466431-47-1.jpg?t=%E6%B1%82%E6%95%B0%E5%88%97%E5%89%8DN%E9%A1%B9%E5%92%8CTn%2CAn+%3D+n+-1%2Cn%E5%B1%9E%E4%BA%8E%E6%AD%A3%E6%95%B4%E6%95%B0+%E6%B1%82Tn+%3D+%28a1%29%5E2+-%28a2%29%5E2%2B%28a3%29%5E2-%28a4%29%5E2.%2B%28-1%29%5E%28n%2B1%29+%2A+bn%5E2)
求数列前N项和Tn,An = n -1,n属于正整数 求Tn = (a1)^2 -(a2)^2+(a3)^2-(a4)^2.+(-1)^(n+1) * bn^2
求数列前N项和Tn,An = n -1,n属于正整数 求Tn = (a1)^2 -(a2)^2+(a3)^2-(a4)^2.+(-1)^(n+1) * bn^2
求数列前N项和Tn,An = n -1,n属于正整数 求Tn = (a1)^2 -(a2)^2+(a3)^2-(a4)^2.+(-1)^(n+1) * bn^2
分N是奇数还是偶数讨论,留意到a1是0,所以-a2^2+a3^2=(a3+a2)*(a3-a2)=a3+a2.以此类推,当n为奇数的时候,Tn=1+2+...+(n-1)=(n-1)*n/2,当n为偶数时Tn=(n-2)*(n-1)/2-(n-1)^2=-(n-1)*n/2.这时发现两个式子可以合并,不再赘述…手机打字,若有不妥,多多包涵
此数列An为等差数列,求和,采用并项求和的方法,即两项合作一项,
若n为奇数,
Tn=0^2-1^2+2^2+……-(n-2)^2+(n-1)^2=-1-5-……-(2n-5)+(n-1)^2=(n^2-n)/2 注:共n项,前n-1项两项合为一项,共(n-1)/2, 且为等差数列,套用等差数列前n项和公式.
若n为偶数,
Tn=0^2-1^2+2^2+……+...
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此数列An为等差数列,求和,采用并项求和的方法,即两项合作一项,
若n为奇数,
Tn=0^2-1^2+2^2+……-(n-2)^2+(n-1)^2=-1-5-……-(2n-5)+(n-1)^2=(n^2-n)/2 注:共n项,前n-1项两项合为一项,共(n-1)/2, 且为等差数列,套用等差数列前n项和公式.
若n为偶数,
Tn=0^2-1^2+2^2+……+(n-2)^2-(n-1)^2=-1-5-……-(2n-1)=(-n^2 +n)/2
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