等比數列前N項和的公式推導

設等比數列的前n項和為S(n), 等比數列的第一項為a1,比值為q。

(1)S(n) = a1 + a1 * q + a1 * q ^ 2 + .... + a1 * q ^ (n - 1);
(2)S(n+1) = a1 + a1 * q + a1 * q ^ 2 + .... + a1 * q ^ (n - 1) + a1 * q ^ n;
由(2)式減(1)式得
(3)S(n+1) - s(n) = a1 * q ^ n;
由S(n) * q 得
(4)S(n) * q = a1 * q + a1 * q ^ 2 + ... + a1 * q ^ n;

由(2)- (4)得
(5)S(n+1) - S(n) * q = a1;

由(3) - (5) 得
[S(n+1) - S(n)] - [S(n+1) - S(n) * q] = a1 * q ^ n - a1;
化簡得:
S(n) * q - S(n) = a1 * (q ^ n - 1);
即:S(n) = a1 * (q ^ n - 1) / (q - 1);

?

轉載于:https://www.cnblogs.com/sdlypyzq/archive/2013/02/21/2920533.html

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