a)2×1+2×[tex] 3^{1} [/tex]+2×3³+...+2×[tex]3^{99}=[/tex]
=2+[tex] 2^{1+3+......+99} [/tex]
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×[tex] 3^{1} [/tex]+2×3³+...+2×[tex]3^{99}=[/tex] devine
=2+[tex] 2^{2500} [/tex]
b) [tex]2^{20}+2^{21}+...+2^{50}= [/tex]
[tex]=2^{20+21+....+50}[/tex]
(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
[tex]2^{20}+2^{21}+...+2^{50}=2^{1085}[/tex]
