Răspuns :
1. x-numar natural
c-cat
r (restul)=c(catul)
x:9= C r.C >>>teorema imparitii cu rest>>> x=9C+C
ii dam valori lui c
c=1 >> x= 9*1+1
c=2 >> x= 9*2+2
c=3 >> x= 9*3+3 ...etc
S= 9*1+1+9*2+2+.......+9*n+n
s= (1+2+...+n) + 9(1+2+3+..+n) // folosim formula sumei lui gaus
s=10 [n(n+1)/2)]
s=5n(n+1)
2. problema nr 2 e similara
c-cat
r (restul)=c(catul)
x:9= C r.C >>>teorema imparitii cu rest>>> x=9C+C
ii dam valori lui c
c=1 >> x= 9*1+1
c=2 >> x= 9*2+2
c=3 >> x= 9*3+3 ...etc
S= 9*1+1+9*2+2+.......+9*n+n
s= (1+2+...+n) + 9(1+2+3+..+n) // folosim formula sumei lui gaus
s=10 [n(n+1)/2)]
s=5n(n+1)
2. problema nr 2 e similara
1) x-numar natural
c-cat
r (restul)⇒c(catul)
x:9= C r.C >>>teorema imparitii cu rest⇒ x=9C+C
ii dam valori lui c
c=1 x= 9*1+1
c=2 ⇒ x= 9*2+2
c=3 ⇒ x= 9*3+3 ...etc
S= 9*1+1+9*2+2+.......+9*n+n
S= (1+2+...+n) + 9(1+2+3+..+n) // folosim formula sumei lui gaus
S=10 [n(n+1)/2)]
S=5n(n+1)
c-cat
r (restul)⇒c(catul)
x:9= C r.C >>>teorema imparitii cu rest⇒ x=9C+C
ii dam valori lui c
c=1 x= 9*1+1
c=2 ⇒ x= 9*2+2
c=3 ⇒ x= 9*3+3 ...etc
S= 9*1+1+9*2+2+.......+9*n+n
S= (1+2+...+n) + 9(1+2+3+..+n) // folosim formula sumei lui gaus
S=10 [n(n+1)/2)]
S=5n(n+1)