(1)求数列{(2n+1).2n}的前n项和 (2)求数列{(n+1)(n+2)/1}的前n项
(1)求数列{(2n+1).2n}的前n项和 (2)求数列{(n+1)(n+2)/1}的前n项和
a(1)+a(2)+a(3)+...+a(n-1)+a(n)=(4/3)[1*2*3-0 + 2*3*4-1*2*3 + 3*4*5-2*3*4 + ... + (n-1)n(n+1)-(n-2)(n-1)n + n(n+1)(n+2)-(n-1)n(n+1)] - [1*2-0 + 2*3-1*2 + 3*4-2*3 + ... + (n-1)n-(n-2)(n-1) + n(n+1)-(n-1)n]
=(4/3)[n(n+1)(n+2)] - [n(n+1)]
=4n(n+1)(n+2)/3 - n(n+1)
=n(n+1)[4n+8 - 3]/3
=n(n+1)(4n+5)
b(n)=1/[(n+1)(n+2)] = 1/(n+1) - 1/(n+2),
b(1)+b(2)+b(3)+...+b(n-1)+b(n)=1/2-1/3 + 1/3-1/4 + 1/5-1/4 + ... + 1/n-1/(n+1) + 1/(n+1)-1/(n+2)
=1/2 - 1/(n+2)
=(n+2-2)/[2(n+2)]
=n/[2(n+2)]
(1)求数列{(2n+1).2n}的前n项和 (2)求数列{(n+1)(n+2)\/1}的前n项
=4n(n+1)(n+2)\/3 - n(n+1)=n(n+1)[4n+8 - 3]\/3 =n(n+1)(4n+5)b(n)=1\/[(n+1)(n+2)] = 1\/(n+1) - 1\/(n+2),b(1)+b(2)+b(3)+...+b(n-1)+b(n)=1\/2-1\/3 + 1\/3-1\/4 + 1\/5-1\/4 + ... + 1\/n-1\/(n+1) + 1\/(n+1)-1\/(n+2...
1.求通项公式为an=(2n+1)\/{n(n+1)(n+2)}的数列前n项和Sn 2.数列{an}...
=2\/(n+1) - 2\/(n+2) + (1\/2)\/[n(n+1)] - (1\/2)\/[(n+1)(n+2)],s(n)=2[1\/2-1\/3 + 1\/3-1\/4 + ... + 1\/(n+1)-1\/(n+2)] +(1\/2)[1\/[1*2] - 1\/[2*3] + 1\/[2*3] - 1\/[3*4] + ... +1\/[n(n+1)] - 1\/[(n+1)(n+2)]=2{1...
求数列{(2n+1)\/【n的平方*(n+1)的平方】}的前n项和
(2n+1)\/[n^2(n+1)^2]=1\/n^2-1\/(n+1)^2 sn=1\/1^2-1\/2^2+1\/2^2-1\/3^2+1\/3^2-1\/4^2+...+1\/n^2-1\/(n+1)^2 =1-1\/(n+1)^2 =(n^2+2n)\/(n+1)^2
Sn=n(n+1)(n+2). (1)求数列{an}的通项公式;(2)求数列{1\/an}的前n...
=3n(n+1)n=1时，a1=3×1×(1+1)=6，同样满足通项公式 数列{an}的通项公式为an=3n(n+1)1\/an=1\/[3n(n+1)]=(1\/3)[1\/n -1\/(n+1)]Tn=1\/a1+1\/a2+...+1\/an =(1\/3)[1\/1-1\/2+1\/2-1\/3+...+1\/n-1\/(n+1)]=(1\/3)[1- 1\/(n+1)]=n\/[3(n+1)]...
数列(2n+1)(2分之1)的n次方的前n项和
利用错位相减法即可求解。令an=(2n+1)\/(2^n)Sn是{an}的前n项和 则Sn=3\/2+5\/4+7\/8+……+(2n+1)\/(2^n) ① 乘2 2Sn=3+5\/2+7\/4+……+(2n+1)\/(2^(n-1)) ② 错位相减 得Sn= 5 -(2n+5)\/(2^n)
求数列{n(n+1)(2n+1)}的前n项和
1^3 + 2^3 +3^3 + ……+ n^3 = [n(n+1)\/2]^2 因此可以把所求式子展开,然后利用上面的2个公式 n(n+1)(2n+1) = (n^2+n)(2n+1) = 2n^3 +3n^2 +n Sn = 2*(1^3+2^3+……+n^3) + 3*(1^2+2^2+ ……+n^2) + (1+2+……+n)= 2*[n(n+1)\/2]^...
求数列Cn=(2n+1)\/(2^n+1)的前n项和(请写个详细的过程,谢谢)
解答：错位相减法 设前n项和是Sn 则 Sn=3\/2²+5\/2³+ 7\/2^4+...+(2n-1)\/2^n+ (2n+1)\/2^(n+1) ① ①乘以1\/2 (1\/2)Sn= 3\/2³+5\/2^4+...+(2n-1)\/2^(n+1)+(2n+1)\/2^(n+2)则 ①-② (1\/2)Sn=3\/4+2【1\/2³+1\/2^4...
求 数列{n(n+1)(2n+1)}的前n项和
方法很常规，n(n+1)(2n+1)=2n*n*n+3n*n+n 再利用立方和 平方和公式，化简；关于平方和与立方和公式的证明，可用数学归纳法或二次项展开法
已知数列{an}的通项公式an=2n+1\/[n(n+1)]^2,求它的前n项和
变通项公式 an=[(n＋1)^2－n^2]\/[n(n＋1)]^2=1\/n^2－1\/(n＋1)^2 a1=1-1\/2^2 a2=1\/2^2-1\/3^2 a3=1\/3^2-1\/4^2 …an=1\/n^2－1\/(n＋1)^2 Sn=1-1\/2^2+1\/2^2-1\/3^2+...+1\/n^2－1\/(n＋1)^2=1－[1\/(n＋1)^2]...
求数列{n(n+1)(n+2)}的前n项和
k+1)(2k+1)＝2k3+3k2+k，∴Sn＝nk＝1k(k+1)(2k+1)=nk＝1(2k3+3k2+k)，将其每一项拆开再重新组合得：Sn＝2nk＝1 k3+3nk＝1 k2+nk＝1 k=2（13+23+…+n3）+3（12+22+…+n2）+（1+2+…+n）=n2(n+1)22+n(n+1)(2n+1)2+n(n+1)2=n(n+1)2(n+2)2．
