Question

CUM REZOLV SUMA?
1+5+9+...+4N+1?

Answer 1

1,5,9,...4n+1

Met1:

Progresie aritmetica de (4n+1 - 1 ) / 4 + 1 = n+1 termeni, n∈N si ratie r=4.

S=(n+1)(1+4n+1)/2=(n+1)(4n+2)/2=2(n+1)(2n+1)/2 = (n+1)(2n+1)

Met2: inductie completa(matematica)

verificam pt k=0: 1, S=1=(0+1)(2x0+1)=1x1=1

k=1: 1+5 = (1+1)(2x1 + 1 ) = 2x3=6 se verifica.

Pp avevarat pt n=k ca 1+5+9+4k+1 = (k+1)(2k+1)   (*)

Vom demonstra ca formula este valabila si pentru n=k+1, adica pasul propriu-zis de inductie completa:

V.D. 1+5+9+...+4k+1  +  4(k+1)+1 = (k+1+1)(2(k+1)+1)

M1(cf ip de inductie)=(k+1)(2k+1) + 4(k+1) + 1 = 2k^2 + 3k + 1 + 4k + 5 = 2k^2 + 7k + 6

M2=(k+2)(2k+3) = 2k^2 + 7k + 6,

⇒ M1=M2, deci presupunerea facuta la (*) este adevarata.

Q.E.D.