Question

要怎么证明这几个公式

Answer 1

(1)
a(n+1) - a(1) = nd
a(n+2) - a(2) = nd
...
a(n+i) - a(i) = nd
S(2n)-S(n) = n^2 d
余者同理.....

(2)
S(2n) = a1 + a(2) + ... + a(n) + a(n+1) + a(n+2) + ... + a(2n)
S(2n) = a(2n) + a(2n-1) + ... + a(n+1) + a(n) + a(n-1) + ... + a(1)
2S(2n) = 2n ( a1 + a(2n) ) = 2n(a(n) + a(n+1))
S(2n) = n(a(n)+a(n+1))

a(2)-a(1) = d
a(4)-a(3) = d
....
a(2n) - a(2n-1) = d
S偶 - S奇 = nd

S奇 = n[a(1) + a(2n-1)]
S偶 = n[a(2) + a(2n)]
S奇/S偶 = (a(1)+a(2n-1))/(a(2)+a(2n)) = a(n)/a(n+1)

S(2n-1) = S(2n) - a(2n) = n a(n) + n a(n+1) - a(2n) = 2n a(n) + nd - a(2n)
= (2n-1) a(n)

S奇 - S偶 = a(2n) - [S(2n)偶 - S(2n)奇] = a(2n) - nd = a(n)

S奇/S偶 = (a(1)+a(2n-1))/(a(2)+a(2n-2)) = na(n)/[(n-1)a(n)] = n/(n-1)