Question

(2n-1)31+345.1.3 + 2.32 + 3.33 ++ n.3-

Answer 1

The given statement is,

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

For n=1,

P( 1 ):1⋅3= ( 2−1 ) 3 1+1 +3 4 P( 1 ):3= ( 2−1 )9+3 4 P( 1 ):3= 9+3 4 P( 1 ):3=3

Thus, P( 1 ) is true.

Substitute n=k in equation (1).

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

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

P( k+1 ): ( 2k−1 ) 3 k+1 +3 4 +( k+1 )⋅ 3 ( k+1 ) = ( 2( k+1 )−1 ) 3 k+1+1 +3 4 P( k+1 ): ( 2k−1 ) 3 k+1 +3+4( k+1 )⋅ 3 k+1 4 = ( 2k+2−1 ) 3 k+2 +3 4 P( k+1 ): ( ( 2k−1 )+4( k+1 ) ) 3 k+1 +3 4 = ( 2k+1 ) 3 k+2 +3 4 P( k+1 ): ( 6k+3 ) 3 k+1 +3 4 = ( 2k+1 ) 3 k+2 +3 4

Further simplify,

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

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

Hence, statement P( n ) is true.