數學問題一問

Question: Prove that 1²+2²+3²......+n² = 1/6 {n(n+1)(2n+1)}
1. When n=1
　LHS = 1² = 1
　RHS = 1/6 (1)(2)(3) = 1
　So, it is true when n = 1
2. Assume it is true when n = k
　 1²+2²+3²......+k² = 1/6 {k(k+1)(2k+1)}

　When n = k+1
　LHS = 1²+2²+3²......+k² +(k+1)²
　　　= 1/6 {k(k+1)(2k+1)} + (k+1)²
　RHS = 1/6 {(k+1)[(k+1)+1][2(k+1)+1]}
　　　= 1/6 {(k+1)(k+2)(2k+3)}
　　　= 1/6 {2k³+9k²+13k+6}
　　　= 1/6 {(2k³+3k²+k)+(6k²+12k+6)}
　　　= 1/6 {k(2k²+3k+1)} + 1/6 {6(k²+2k+1)}
　　　= 1/6 {k(k+1)(2k+1)} + (k+1)²
　LHS = RHS
　It is ture for n = k+1
Therefore, 1²+2²+3²......+n² = 1/6 {n(n+1)(2n+1)} is ture,
for all positive integer.

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