Question

$Fie~sirul~(a_{n}),a_{0}=0,a_{1}=1,a_{2}=4,a_{k+1}=2a_{k}-a_{k-2}-2k+5. \\ Sa ~se~determine~a_{n}~si~suma~primilor~n~termeni~ai~sirului.$
Multumesc!

Answer 1

k=2 a3=2a2-ao-2*2+5=2*4-4+5=9=3²
k=3  a4=2a3-a1-2*3+5=2*9-1-6+5=16=4²
............................................................
k=n    an+1=(n+1)² egalitatea  se  demonstreaza  prin  inductie.
Presupunem an+1 adevarata .an+1=> an+2
an+1=(n+1)²
k=n+1 an+2=(n+2)²
an+2=2an+1-an-1-2*(n+1)+5=
2*(n+1)²-`(n-1)²-2n-2+5
2n²+4n+2-n²+2n-1-2n-2+5=
n²+4n+4=(n+2)²
an+1=>an+2 Egalitatea  este  demonstrata
a0+a1+a2+...+an-1=
0+1+4+9+...(n-1)²=suma  primeloer  (n-1)  nr  naturale