Question

If , then prove where n is any positive integer

Answer 1

Using the mathematical induction formula,

Step 1:

Consider the mathematical induction for the n positive integer is P( n ) if,

A=[ 3 −4 1 −1 ]

Then,

A n ︸ LHS = [ 1+2n −4n n 1−2n ] ︸ RHS      ( n∈N ) (1)

Step 2:

Put n=1 in the left hand side (LHS) of the equation (1).

LHS= A 1 =A =[ 3 −4 1 −1 ]

And, put n=1 in the right side of the equation (1)

RHS=[ 1+2( 1 ) −4( 1 ) 1 1−2( 1 ) ] =[ 3 −4 1 −1 ] =LHS

Here,

LHS=RHS

Therefore, P( n ) is true for n=1 .

Step 3:

Consider P( k ) to be true.

A=[ 3 −4 1 −1 ]

Then,

A k ︸ LHS = [ 1+2k −4k k 1−2k ] ︸ RHS      ( k∈N ) (2)

Step 4:

Proving that P(k + 1) is true.

Replace k with k+1 in the equation (2).

P( k+1 ): if A=[ 3 −4 1 −1 ]

Then,

A k+1 =[ 1+2(k+1) −4(k+1) k+1 1−2(k+1) ]

Take LHS,

A k+1 = A k A =[ 1+2k −4k k 1−2k ][ 3 −4 1 −1 ] =[ ( 1+2k )3−4k( 1 ) ( 1+2k )( −4 )−4k( −1 ) k( 3 )+( 1−2k )1 k( −4 )+( 1−2k )( −1 ) ]

Simplify the above expression,

A k+1 =[ 1+2( k+1 ) −4( k+1 ) 1+k 1−2( k+1 ) ] =RHS

Prove that P( k+1 ) is true.

Hence, prove that P( n ) is true for all n∈N