Question

已知数列{an}是等差数列,前n项和为Sn,a3=6,s3=12,求S1+S2+S3+S4+......+Sn的值

Answer 1

解：因为a3=6,s3=12,所以a1+a2=6,设首项为a1,公差为d,则a1+2d=6, 且2a1+d=6,解得a1=2,d=2,
所以an=a1+(n-1)d=2+(n-1)*2=2n, sn=(2+2n)*n/2=n(n+1),
s1+s2+s3+....+sn=1*2+2*3+3*4+....+n(n+1)

Answer 2

s3=6-2d+6-d+6=12
∴d=2
∴a1=2
∴sn=n(n+1)=n²+n
∴S1+S2+S3+S4+......+Sn=n(n+1)(2n+1)/6+n(n+1)/2
=n(n+1)(n+2)/3

Answer 3

a3=a1+2d=6
s3=a1+a2+a3=3a1+3d=12
a1= 2 d=2
an=a1+(n-1)d=2n
sn=n*(a1+an)/2=n(n+1)/2=n^2/2+n/2
S1+S2+S3+S4+......+Sn
=1/2*n(n+1)(n+2)/6+1/2 *n(n+1)/2
=n(n+1)(n+2)/12+n(n+1)/4