Question

Aratati ca:a) 2+4+6+...+100 nu este patrat perfect. b) nr.1+3+5+...+199 este patrat perfect.

Answer 1

a) 
2(1+2+3+...+50)
= 2*50*51:2=50*51- nu este patrat perfect
b)
nr termenilor= (199-1):2+1=198:2+1=99+1=100
suma=(199+1)*100:2=200*100:2=100*100=100²- patrat perfect

Answer 2

$\displaystyle a).2+4+6+...+100=2(1+2+3+...+50)=\not 2 \cdot \frac{50(50+1)}{\not 2} = \\ \\ =50 \cdot 51=2550 \not = p.p. \\ \\ b).1+3+5+...+199 \\ \\ 199=1+(n-1) \cdot 2 \Rightarrow 199=1+2n-2 \Rightarrow 2n=199-1+2 \Rightarrow \\ \\ \Rightarrow 2n=200 \Rightarrow n= \frac{200}{2} \Rightarrow n=100 \\ \\ S_{100}= \frac{2 \cdot 1+(100-1) \cdot 2}{2} \cdot 100 \\ \\ S_{100}=(2+99 \cdot 2) \cdot 50 \\ \\ S_{100}=(2+198) \cdot 50 \\ \\ S_{100}=200 \cdot 50\\ \\ S_{100}=10000 \\ \\ S_{100}=100^2=p.p.$