數學歸納法証明

1.

Let S (n) be the statement that n(n+1)(2n+1) is divisible by 3

When n = 1,

(1)(2)(3) = 6 = 3 ( 2 )

therefore S(1) is true

Assume that S(k) is true for some integers k.
i.e. k( k + 1 )( 2k + 1 ) = 3M , where M is an integer

When n = k + 1

( k + 1 ) ( k + 2 ) ( 2k + 3 ) = ( k + 1 ) ( 2k^2 + 7k + 6 )
= ( k + 1 ) ( 2k^2 + k + 6k + 6 )
= ( k + 1 ) ( k ) ( 2k + 1 ) + ( k + 1 ) ( 6k + 6 )
= 3M + 6 ( k + 1 )( k + 1 )
= 3 [ M + 2 ( k + 1 ) ( k + 1 ) ]
therefore S ( k + 1 ) is also true

By induction, S (n) is true for all n.

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2.

Let S (n) be the statement that 2²+4²+6²+...+(2n)²=[2n(2n+1)(n+1)]/3

When n =1 ,
LHS = 2²
= 4
= 2(3)(2) / 3
= RHS

therefore S(1) is true

Assume that S (k) is true
i.e.
2²+4²+6²+...+(2k)²=[2k(2k+1)(k+1)]/3

When n = k + 1,
2²+4²+6²+...+(2k)² + (2k+2)² = [ 2k(2k+1)(k+1)]/3 + (2k+2)²
= [ 2k(2k+1)(k+1) + 3 (2k+2)² ] / 3
= { 2( k + 1 ) [ k(2k+1) + 6( k +1 )] } / 3
= { 2( k + 1 ) [ 2k^2 + k + 6k + 6 ] } / 3
= [2(k+1)(2k+3)(k+2)]/3
therefore S( k + 1) is true

By induction, S (n) is true for all n.