1.2+2.2^(2)+3.2!.4+n.2"=(n-1)2^(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

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

Prove, by method of induction, for all n in N : 2 + 3.2 +4.2^2 +...+ (n + 1) (2^(n-1)) = n.2^n

1*2+2*2^(2)+3*2^(3)+*2^(n)=(n-1)2^(n+1)+2

1+2.2+3.2^(2)+4.2^(3)+...+n*2^(n-1) = (i) 1+ (1+n) 2^(n) (ii) 1- (1+n) 2^(n) (iii) 1- (1-n) 2^(n) (iv) 1+ (1-n) 2^(n)

[ If 2 is the sum of infinity of a G.P.,whose first clement is 1 ,then the sum of the first n terms is [ 1) (2^(n)-1)/(2^(n)), 2) (2^(n)-1)/(2^(n-1)), 3) (2^(n-1)-2)/(2), 4) (2^(n-1)-1)/(2^(n))]]

1 ^(2) + 2^(2) + 3^(2) + . . . + n^(2) = (n (n + 1) (2 n + 1))/( 6)

Prove by mathematical induction that, 1^(2)-2^(2)+3^(2)-4^(2)+ . . .. +(-1)^(n-1)*n^(2)=(-1)^(n-1)*(n(n+1))/(2),ninNN .