在等差数列{an}中,公差d=2分之一,前100项的和S100=45,则a1+a3+a5+…+a在等差数列{an}中,公差d=2分之一,前100项的和S100=45,则a1+a3+a5+…+a99=为什么a2加a4加a6..减a1加a3加a5...等于50d呢?

Question

在等差数列{an}中,公差d=2分之一 ,前100项的和S100=45,则a1+a3+a5+…+a
在等差数列{an}中,公差d=2分之一
,前100项的和S100=45,则a1+a3+a5+…+a99= 为什么a2加a4加a6..减a1加a3加a5...等于50d呢?

Answer 1

a(n) = a + (n-1)/2,
s(n) = na + n(n-1)/4.
45 = s(100) = 100a + 25*99 = 100a + 55*45,
a = -45*54/100 = -9*27/10 = -243/10,
a(n) = -243/10 + (n-1)/2.
a(2n-1) = -243/10 + (2n-2)/2 = -243/10 + (n-1),
a(1)+a(3)+...+a(2*n-1) = -243n/10 + n(n-1)/2,
a(1)+a(3)+...+a(2*50-1) = -243*50/10 + 50*49/2 = -243*5 + 25*49 = 5(5*49 - 243)
= 5(5*50 - 5 - 243)
= 5(250 - 248)
= 10
a(n) = a + (n-1)d,
a(2n) = a + (2n-1)d,
a(2n-1) = a + (2n-2)d.
a(2n) - a(2n-1) = d.
[a(2)+a(4)+...+a(2*50)] - [a(1)+a(3)+...+a(2*50-1)] = [a(2)-a(1)] + [a(4)-a(3)] + ...+ [a(2*50)-a(2*50-1)] = 50*d = 50d.