Respuesta :
Answer:
the first partial sum [tex]\mathbf{S_1= \dfrac{1}{2}}[/tex]
the second partial sum [tex]\mathbf{S_2= \dfrac{3}{4} }[/tex]
the third partial sum [tex]\mathbf{S_3= \dfrac{7}{8}}[/tex]
the fourth partial sum [tex]\mathbf{S_4= \dfrac{15}{16}}[/tex]
the fifth partial sum [tex]\mathbf{S_5= \dfrac{31}{32}}[/tex]
the sixth partial sum [tex]\mathbf{S_6= \dfrac{63}{64}}[/tex]
Step-by-step explanation:
The term of the sequence are given as : [tex]\dfrac{1}{2}[/tex], [tex]\dfrac{1}{2^2}[/tex], [tex]\dfrac{1}{2^3}[/tex], [tex]\dfrac{1}{2^4 }[/tex] , . . .
The nth term for this sequence is , [tex]\mathtt{a_n =( \dfrac{1}{2})^n}[/tex]
The nth partial sum of the sequence for [tex]\mathtt{a_1,a_2,a_3.... a_n}[/tex] is [tex]\mathtt{S_n}[/tex]
where;
[tex]\mathtt{S_n = a_1 +a_2+a_3+ ...+a_n}[/tex]
The first partial sum is: [tex]\mathtt{S_1= a_1}[/tex]
[tex]\mathtt{S_1= (\dfrac{1}{2})^1}[/tex]
[tex]\mathbf{S_1= \dfrac{1}{2}}[/tex]
Therefore, the first partial sum [tex]\mathbf{S_1= \dfrac{1}{2}}[/tex]
The second partial sum is: [tex]\mathtt{S_2= a_1+a_2}[/tex]
[tex]\mathtt{S_2= (\dfrac{1}{2})^1 + (\dfrac{1}{2})^2}[/tex]
[tex]\mathtt{S_2= \dfrac{1}{2} + \dfrac{1}{4}}[/tex]
[tex]\mathtt{S_2= \dfrac{2+1}{4} }[/tex]
[tex]\mathbf{S_2= \dfrac{3}{4} }[/tex]
Therefore, the second partial sum [tex]\mathbf{S_2= \dfrac{3}{4} }[/tex]
The third partial sum is : [tex]\mathtt{S_3= a_1+a_2+a_3}[/tex]
[tex]\mathtt{S_3= (\dfrac{1}{2})^1 + (\dfrac{1}{2})^2+(\dfrac{1}{2})^3 }[/tex]
[tex]\mathtt{S_3= \dfrac{1}{2} + \dfrac{1}{4}+\dfrac{1}{8}}[/tex]
[tex]\mathtt{S_3= \dfrac{4+2+1}{8}}[/tex]
[tex]\mathbf{S_3= \dfrac{7}{8}}[/tex]
Therefore, the third partial sum [tex]\mathbf{S_3= \dfrac{7}{8}}[/tex]
The fourth partial sum : [tex]\mathtt{S_4= a_1+a_2+a_3+a_4}[/tex]
[tex]\mathtt{S_4= (\dfrac{1}{2})^1 + (\dfrac{1}{2})^2+(\dfrac{1}{2})^3+(\dfrac{1}{2})^4 }[/tex]
[tex]\mathtt{S_4= \dfrac{1}{2} + \dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}}[/tex]
[tex]\mathtt{S_4= \dfrac{8+4+2+1}{16}}[/tex]
[tex]\mathbf{S_4= \dfrac{15}{16}}[/tex]
Therefore, the fourth partial sum [tex]\mathbf{S_4= \dfrac{15}{16}}[/tex]
The fifth partial sum : [tex]\mathtt{S_5= a_1+a_2+a_3+a_4+a_5}[/tex]
[tex]\mathtt{S_5= (\dfrac{1}{2})^1 + (\dfrac{1}{2})^2+(\dfrac{1}{2})^3+(\dfrac{1}{2})^4 +(\dfrac{1}{2})^5 }[/tex]
[tex]\mathtt{S_5= \dfrac{1}{2} + \dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}}[/tex]
[tex]\mathtt{S_5= \dfrac{16+8+4+2+1}{32}}[/tex]
[tex]\mathbf{S_5= \dfrac{31}{32}}[/tex]
Therefore, the fifth partial sum [tex]\mathbf{S_5= \dfrac{31}{32}}[/tex]
The sixth partial sum: [tex]\mathtt{S_5= a_1+a_2+a_3+a_4+a_5+a_6}[/tex]
[tex]\mathtt{S_6= (\dfrac{1}{2})^1 + (\dfrac{1}{2})^2+(\dfrac{1}{2})^3+(\dfrac{1}{2})^4 +(\dfrac{1}{2})^5 +(\dfrac{1}{2})^6 }[/tex]
[tex]\mathtt{S_6= \dfrac{1}{2} + \dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64} }[/tex]
[tex]\mathtt{S_6= \dfrac{32+16+8+4+2+1}{64}}[/tex]
[tex]\mathbf{S_6= \dfrac{63}{64}}[/tex]
Therefore, the sixth partial sum [tex]\mathbf{S_6= \dfrac{63}{64}}[/tex]
