Question

Demonstreaza ca numarul A este patrat perfect , unde A egal cu 1 plus 2 plus 3 plus ....plus 100 plus 51 ori 101

Answer 1

A=1+2+3+4+5+.....+100+51*101
A=100*101/2+51*101
A=50*101+51*101
A=101(50+51)
A=101²

Answer 2

[tex]A=1+2+3+...+100+(51*101)\\ Observatie!Avem~o~Suma~Gauss,vom~aplica~formula~sumei\\ Gauss:[n(n+1)]:2\\ Copiem~exercitiul...\\ 1+2+3+...+100+5151\\ aplicam~Gauss:(100*101):2+(51*101)=>50*101+51*101=>10201 \\ Observatie!Ultima~cifra~a~oricarui~patrat~perfect~este\\ mereu~una~din~cifrele: 0, 1, 4, 5, 6, 9.Cum~numarul~nostru\\ se~termina~in~1~rezulta~A=patrat~perfect.\\\\ [/tex]