Question

aratati ca nr. N=(1+2+3+....+99)·2+100 este patrat perfect

Answer 1

$\boxed{1+2+3+...+n = \dfrac{n\cdot(n+1)}{2}}\rightarrow Suma~lui~Gauss.\\ \\\\ N = (1+2+3+...+99)\cdot 2+100 \\ \\ N = \dfrac{99\cdot(99+1)}{2}\cdot 2+100 \\ \\ N = 99\cdot (99+1)+100 \\ \\ N = 99\cdot 100+100 \\ \\ N = 100\cdot (99+1) \\ \\ N = 100\cdot 100 \\ \\ N = 100^2 \rightarrow patrat~ perfect.$

Answer 2

$N=(1+2+3+...+99)*2+100 \\ 99*100/2=9900/2=4950 \\ N=4950*2+100 \\ N=9900+100=10000 \\ \\ \sqrt{10000}=100$