Question

23+234 )0.3,4.4

Answer 1

The given statement is,

P( n ):1⋅2⋅3+2⋅3⋅4+...+n( n+1 )( n+2 )= n( n+1 )( n+2 )( n+3 ) 4 (1)

For n=1,

P( 1 ):1⋅2⋅3= 1( 1+1 )( 1+2 )( 1+3 ) 4 P( 1 ):1⋅2⋅3=1⋅2⋅3

Thus, P( 1 ) is true.

Substitute n=k in equation (1).

P( k ):1⋅2⋅3+2⋅3⋅4+...+k( k+1 )( k+2 )= k( k+1 )( k+2 )( k+3 ) 4 (2)

According to the principle of mathematical induction, assume that the statement P( n ) is true for n=k. If it is also true for n=k+1, then P( n ) is true for all natural numbers.

Substitute n=k+1 in equation (1).

P( k+1 ):1⋅2⋅3+2⋅3⋅4+...+( k+1 )( k+2 )( k+3 ) = ( k+1 )( k+2 )( k+3 )( k+4 ) 4 (3)

Substitute the values from equation (2) into equation (3).

P( k+1 ): k( k+1 )( k+2 )( k+3 ) 4 +( k+1 )( k+2 )( k+3 )= ( k+1 )( k+2 )( k+3 )( k+4 ) 4 P( k+1 ): k( k+1 )( k+2 )( k+3 )+4( k+1 )( k+2 )( k+3 ) 4 = ( k+1 )( k+2 )( k+3 )( k+4 ) 4 P( k+1 ): ( k+1 )( k+2 )( k+3 )( k+4 ) 4 = ( k+1 )( k+2 )( k+3 )( k+4 ) 4

It is proved that P( k+1 ) is true whenever P( k ) is true.

Hence, statement P( n ) is true.