Question

Calculati:
a. (2 x 1 + 1) + ( 2 x 2 + 1) + ... + (2 x n +1 )
b. (2 x 2 - 3 ) + ( 2 x 3 - 3) + ... + ( 2 x n - 3 )

Answer 1

a. (2 x 1 + 1) + ( 2 x 2 + 1) + ... + (2 x n +1 ) =
=3+5+....+(2n+1)=
adunam si scadem 1+2+4+....+2n 
1+2+4+....+2n =1+2(1+2+....+n)=1+2n(n+1)/2=1+n(n+1)
1+2+3+4+....+2n+1-[1+n(n+1)]=
=(2n+1)(2n+1+1)/2-[1+n(n+1)]=
=(2n+1)(2n+2)/2-1-n(n+1)=
=(2n+1)(n+1)-1-n(n+1)=
=(n+1)(2n+1-n)-1=
=(n+1)*(n+1)-1=
=(n+1)²-1

b. (2 x 2 - 3 ) + ( 2 x 3 - 3) + ... + ( 2 x n - 3 )=
=(4-3)+(6-3)+....+(2n-3)=
=1+3+5....+(2n-3)=
adunam si scadem 2+4+....+2n-2 
2+4+....+2n-2 =2(1+2+....+(n-1)=2(n-1)(n-1+1)/2=(n-1)*n
=1+2+3+4+5....+(2n-2)+(2n-3)- (n-1)*n=(2n-3) (2n-3+1)/2-(n-1)*n=
= (2n-3) (2n-2)/2-(n-1)*n=(2n-3) (n-1)-(n-1)*n=(n-1)(2n-3-n)=(n-1)(n-3)