Question

S=1+5+9+.........+(4n+1) sa fac cu inductia matematica

Answer 1

4(1+2+...+n) +n+1=4n(n+1)/2+n+1=2n(n+1) +(n+1)= (n+1)(2n+1)
 verificam pt n=1
1+5=(1+1)*(2*1+1)=2*3=6 adevarat
 Presupunem adevarat pt n
Sn=1+5+9+.........+(4n+1)=(n+1) (2n+1)
 Studiem pt n+1

 
Sn+1=1+5+9+.........+(4n+1)+ 4(n+1) +1=
             (n+1)(2n+1) +4n+5=
             2n²+3n+1+4n+5=
             2n²+7n+6=2(n+2)(n+3/2)
 am descompus cf formulei a(x-x1)(x-x2) dupa ce am aflat 'radacinile " n1=-2 si n2=-3/2


2(n+2)(n+3/2)=2(n+2)(2n+3)/2= (n+2)(2n+3)= ((n+1)+1) ( 2(n+1)+1)
deci Pn⇒Pn+1, relatia gasita s-a verificat prin inductie, deci este adevarata


Sn=1+5+9+.........+(4n+1)= (n+1)(2n+1)