Given:
[tex]\sum_{n\mathop{=}2}^812(0.4)^{n+1}[/tex]
Required:
Sum of the numbers
Explanation:
Let
[tex]A_n=\sum_{n\mathop{=}2}^812(0.4)^{n+1}[/tex]
when n = 2, Aₙ becomes
[tex]A_2=12(0.4)^{2+1}=12\times(0.4)^3=0.768[/tex]
when n = 3, Aₙ becomes
[tex]A_3=12(0.4)^{3+1}=12\times(0.4)^4=0.3072[/tex]
when n = 4, Aₙ becomes
[tex]A_4=12(0.4)^{4+1}=12\times0.4^5=0.12288[/tex]
when n = 5, Aₙ becomes
[tex]A_5=12(0.4)^{5+1}=12\times0.4^6=0.049152[/tex]
when n = 6, Aₙ becomes
[tex]A_6=12(0.4)^{6+1}=12\times0.4^7=0.0196608[/tex]
when n = 7, Aₙ becomes
[tex]A_7=12(0.4)^{7+1}=12\times0.4^8=0.007866432[/tex]
when n = 8, Aₙ becomes
[tex]A_8=12(0.4)^{8+1}=12\times0.4^9=0.003145728[/tex]
So now,
[tex]\begin{gathered} A=A_1+A_2+A_3+A_4+A_5+A_6+A_7+A_8 \\ \\ A=0.768+0.3072+0.12288+0.049152+0.0196608+0.00786432+0.003145728 \\ \\ A=1.277902848\approx1.28 \end{gathered}[/tex]
Final answer:
The
