数列{an}满足a1=1,an+1=an+4n,则数列{an}的前n项和Sn= 要过程的~~

显示全部楼层 · 2011-10-4 21:37:59

a=an+4n,∴a+2n^2+2n=an+2(n-1)^2+2(n-1)=……=a1=1,∴an=-2(n-1)^2-2(n-1)+1=-2n^2+4n-2-2n+2+1=-2n^2+2n+1,用公式：1^2+2^2+……+n^2=n(n+1)(2n+1)/6,∴Sn=-n(n+1)(2n+1)/3+n(n+2)....

千问 · 2011-10-4 21:37:59

a(n+1)=a(n)+4n=a(n)+2[n(n+1)-(n-1)n],a(n+1)-2n(n+1)=a(n)-2(n-1)n=...=a(1)-0=1,a(n)=2(n-1)n+1=2[(n-1)n(n+1)-(n-2)(n-1)n]/3 + [(n)-(n-1)],s(n)=(2/3)[0-0 + 1*2*3-0 + 2*3*4-1*2*3...

千问 · 2011-10-4 21:37:59

a(n+1)=an+4na(n+1)-an=4na(n)-a(n-1)=4(n-1)……a2-a1=4连加 an-a1=4(n-1)+4(n-2)+4(n-3)+……+4=2n*(n-1)an=2n*(n-1)+a1=2n*(n-1)+1=2n^2-2n+1公式∑n=n*(n+1)/2∑n^2=n*(n+1)*(2n+1)/6...

千问 · 2011-10-4 21:37:59

a(n+1)=a(n)+4n ==> a(n+1)-2n(n+1)=a(n)-2(n-1)n=......=a(1)-2*(1-1)*1=a(1)=1所以a(n)=1+2(n-1)n;所以S(n)=∑a(k)=n+∑2k(k-1)=(2*n^3+n)/3...