Question

Calculează:
a)2x1+2x3^1+2x3^3+...+2x3^99
b)2^20+2^21+...+2^50

Answer 1

a)2×1+2×$3^{1}$+2×3³+...+2×$3^{99}=$

=2+$2^{1+3+......+99}$

1+3+...+99=.... Este de forma 1+3+5+...+2n-1 =n×n;
99=2n -1 de unde 2n = 99+1=100 ⇒ n=50 .Deci suma S=50×50=2500
1+3+...+99=2500

2×1+2×$3^{1}$+2×3³+...+2×$3^{99}=$ devine

=2+$2^{2500}$

b)  $2^{20}+2^{21}+...+2^{50}=$
 
$=2^{20+21+....+50}$

(20+21+....+50)-progresie aritmetica cu ratia r=1, an=50 si a1=20

an=a1+(n−1)×r⇒50 = 20 +(n-1)×1⇒50 = 20+n-1⇒50 - 20 +1 = n - nr de termeni
deci n = 31 de termeni are progresia aritmetica

Sn=n×(a1+an)/2=31×(50+20)/2=1085

  $2^{20}+2^{21}+...+2^{50}=2^{1085}$