Mathematical Induction?

(a)
n = 1 時,
左 = 1
右 = (r^2-2r+1)/(r-1)^2 = 1
故公式成立.

設 n = k 時公式成立, 則,
n = k+1 時,
  1 + 2r + ... + nr^(n-1)
     = 1 + 2r + ... + kr^(k-1) + (k+1)r^k
     = [kr^(k+1)-(k+1)r^k+1]/(r-1)^2 + (k+1)r^k
     = {[kr^(k+1)-(k+1)r^k+1]+(k+1)r^k(r-1)^2}/(r-1)^2
     = {[kr^(k+1)-(k+1)r^k+1]+(k+1)(r^(k+2)-2r^(k+1)+r^k)}/(r-1)^2
     = {(k+1)r^(k+2)-(k+2)r^(k+1)+1}/(r-1)^2
     = [nr^(n+1)-(n+1)r^n+1]/(r-1)^2, 當 r≠1.

故, 對任意正整數 n, 公式成立.

(b)
設 r = 2, n = 15, 則得:
1 + 2×3 + 3×3^2 + ... + 15×3^14
    = (15×3^16-16×3^15+1)/(3-1)^2
    = (29×3^15+1)/4 = 41612304/4 = 10403076

(c)
把 1+r+...+r^(n-1) = (r^n-1)/(r-1) 加到 (a) 之公式:
(1+r+...+r^(n-1)) + [1+3r+...+(2n-1)r^(n-1)]
    = 2+4r+...+2nr^(n-1)
    = 2(1+2r+...+nr^(n-1))
    = 2[nr^(n+1)-(n+1)r^n+1]/(r-1)^2
故
1+3r+...+(2n-1)r^(n-1)
    = 2[nr^(n+1)-(n+1)r^n+1]/(r-1)^2 - (r^n-1)/(r-1)
    = {2[nr^(n+1)-(n+1)r^n+1] - (r^n-1)(r-1)}/(r-1)^2
    = [(2n-1)r^(n+1)-(2n+1)r^n+r+1]/(r-1)^2
取 r = 3, 得:
1 + 3×3 + 5×3^2 + ... + (2n-1)3^(n-1)
    = [(2n-1)3^(n+1) - (2n+1)3^n + 4]/4
    = [(4n-4)3^n+4]/4
    = (n-1)3^n + 1