已知S(1)、S(2),证(r+1)S(2)=(r-1)S(1)^2+2aS(1).S(1)=a+ar+ar^2+.+ar^(n-1)S(2)=a^2+a^2r^2+a^2r^4+.+a^2r^2(n-1)证(r+1)S(2)=(r-1)S(1)^2+2aS(1).

Question

已知S(1)、S(2),证(r+1)S(2) = (r-1)S(1)^2 + 2aS(1).
S(1) = a + ar +ar^2 + .+ ar^(n-1)
S(2) = a^2 + a^2 r^2 + a^2 r^4 +.+ a^2 r^2(n-1)
证(r+1)S(2) = (r-1)S(1)^2 + 2aS(1).

Answer 1

两个都是等比数列,用等比公式求和.
S(1) = a*(r^n - 1) / (r-1)
S(2) = a^2 * (r^2n - 1) / (r^2 - 1)
则 (r-1)S(1)^2 + 2aS(1) = a^2 * (r^n - 1)^2 / (r-1) + 2a^2 * (r^n -1) / (r-1)
= a^2 *[ (r^n - 1)^2 + 2 (r^n -1) ] / (r-1)
= a^2 *[ r^2n -1 ] / (r-1)
(r+1)S(2) = (r+1) a^2 * (r^2n - 1) / (r^2 - 1) = a^2 *[ r^2n -1 ] / (r-1)
所以(r+1)S(2) = (r-1)S(1)^2 + 2aS(1)成立