Form 4 A.MATH M.I

1x2+2x3+3x4+....+n(n +1)= n (n+1)(n+2)/3
當n = 1
LHS = 1x2 = 2
RHS = n(n+1)(n+2)/3 = 1(1+1)(1+2)/3 = 2
所以當 n = 1 時成立
設 n = k 時成立，則
1x2+2x3+3x4+....+k(k+1)= k(k+1)(k+2)/3
當 n = k+1
則 LHS =
1x2+2x3+3x4+....+k(k+1)+(k+1)(k+2)
= k(k+1)(k+2)/3+(k+1)(k+2)
= k(k+1)(k+2)/3+3(k+1)(k+2)/3
= (k+1)(k+2)(k+3)/3
= RHS
結果成立

1x4+2x7+3x10+...+n(3 n+1)=n(n+1)2
當 n = 1
LHS = 1x4 = 4
RHS = n(n+1)2 = 1(1+1)2 = 4
所以當 n = 1 時成立
設 n = k 時成立，則
1x4+2x7+3x10+...+k(3 k+1)=k(k+1)2
當 n = k + 1
LHS =
1x4+2x7+3x10+...+k(3 k+1) + (k+1)(3k+4)
= k(k+1)2 + (k+1)(3k+4)
= (k+1)[k(k+1) + 3k+4]
= (k+1)[k2+ k + 3k+4]
= (k+1)[k2+ 4k+4]
= (k+1)(k+2)2
= RHS
結果成立

13+33+53+...+(2n-1)3=n2(2n2-1)
當 n = 1
LHS = 13 = 1
RHS = n2(2n2-1) = 12(2(1)2-1) = 1
所以當 n = 1 時成立
設 n = k 時成立，則
13+33+53+...+(2k-1)3=k2(2k2-1)
當 n = k + 1
LHS =
13+33+53+...+(2k-1)3 + (2k+1)3
= k2(2k2-1) + (2k+1)3
= 2k4 – k2 + 8k3 + 12k2 +6k + 1
= 2k4 + 8k3 + 11k2 +6k + 1
= (k+1)(2k3+6k2+5k+1)
= (k+1)(2k3+6k2+5k+1)
= (k+1)2(2k2+4k+1)
= (k+1)2(2k2+4k+2-1)
= (k+1)2[2(k+1)2-1)
= RHS
結果成立

1. 3n^3+9n^2+2n
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