Question

1. Să se calculeze:
a) 1+3+5+ ... +27 ; b) 1+5+9+...+37; c) 7+17+27 +...+97;
d) (2^10 – 1)(2^9 - 1)....(2^-9 – 1)(2^-10 – 1).

Answer 1

$\it\ \div\\ \\ a)\ \ S=\dfrac{(1+27)\cdot 14}{2}=28\cdot7=196\\ \\ sau:\\ \\ a)\ \ S=14^2=196\\ \\ \\b)\ \ S=\dfrac{(1+37)\cdot10}{2}=38\cdot5=190\\ \\ \\ c)\ \ S=\dfrac{(7+97)\cdot10}{2}=104\cdot5=520\\ \\ \rule{180}{0.4}\\ \\ d) \ \ P=0,\ \ deoarece\ un\ factor\ este:\ \ 10^0-1=1-1=0$