Question

根据下列条件,求等差数列｛an}的前n项和sn（1）a1=1,an=19,n=10(2)a1=100,d=-5,n=20

Answer 1

a(n) = a + (n-1)d.s(n) = na + (n-1)nd/2.
(1)
a(n) = 1 + (n-1)d.
19 = a(10) = 1 + 9d,d = 2.
a(n) = 1 + 2(n-1) = 2n-1.
s(n) = n + n(n-1) = n^2.
s(10) = 10^2 = 100.
a(10) = 19.
(2)
a(n) = 100 - 5(n-1),
s(n) = 100n - 5(n-1)n/2.
a(20) = 100 - 5*19 = 5*20 - 5*19 = 5
s(20) = 100*20 - 5*19*10 = 10*5*40 - 10*5*19 = 50(40-19) = 50*21 = 1050.
(3)
a(n) = 10 + (n-1)/2,
s(n) = 10n + n(n-1)/4.
a(20) = 10 + 19/2 = 39/2.
s(20) = 10*20 + 20*19/4 = 5*40 + 5*19 = 5*59 = 5*60 - 5 = 295.