Question

数列求解！！急急急！

｛an｝为等差数列，d=1/110，S5=a，S22=b，求S27

Answer 1

S27 = S5 + S22 + d × (27×26 - 22×21 - 5×4) / 2
= a + b + (1/110) × 110
= a + b + 1
答：S27 = a + b + 1
PS，比较一下它们的求和公式就知道怎么做了，很简单，呵呵

Answer 2

由 Sn=na1+d*n(n-1)/2得 S5=5a1+10d=a S22=22a1+231d=b
则a+b=27a1+241d
S27=27a1+251d=27a1+241d+110d=a+b+110*1/110=a+b+1