[3.2^(2)+3^(2)*2^(3)+3^(3)*2^(4)+...+3^(n)*2^(n+1)],[=(12)/(5)(6^(k)-1)]

Using Mathematical induction prove that 3.2^(2)+3^(2).2^(3)+3^(3).2^(4)+…………+3^(n).2^(n+1)=12/6(6^(n)-1) , for all n in N

1^(3)+2^(3)+3^(3)+...+n^(3)=n^(2)((n+1)^(2))/(4)

1^(3)+2^(3)+3^(3)+.....+n^(3)=(n(n+1)^(2))/(4), n in N

If 2^(3)+4^(3)+6^(3)+.. .+(2 n)^(3)=k n^(2)(n+1)^(2) then k=

1.2+2.2^(2)+3.2^(3)dots...n.2^(n)=(n-1)2^(n+1)+2

(1^(3)+2^(3)+...+n^(3))/(1+3+5+...+(2n-1))=((n+1)^( 2))/(4)

Match the following . {:(,"ColumnI",,"ColumnII"),((i) ,1^(2) +2^(2) +3^(2) +....+n^(2) ,(a) ,[(n(n+1))/(2)]^(2)),((ii) , 1^(3) +2^(3) +3^(3) +...+n^(3) ,(b), n(n+1)),((iii),2+4+6+...+2n,( c),(n(n+1)(2n+1))/(6)),((iv),1+2+3+...+n,(d),(n(n+1))/(2)):}

1.2+2.2^(2)+3.2^(3)+....+n.2^(n)=(n-1)2^(n-1)+2

1.2+2.2^(2)+3.2^(3)+....+n.2^(n)=(n-1)2^(n-1)+2

The sum of the series 1+4+3+6+5+8+ upto n term when n is an even number (n^(2)+n)/(4) 2.(n^(2)+3n)/(2) 3.(n^(2)+1)/(4) 4.(n(n-1))/(4)(n^(2)+3n)/(4)