Question

Help please, what is sum of the geometric series E6(4)k-1

Answer 1

B

Explanation:

Edge2021

Answer 2

2,046

Explanation:

`\sum\limits_(k=1)^(5)6\left(4\right)^(k-1)=6\left(4\right)^(1-1)+6\left(4\right)^(2-1)+6\left(4\right)^(3-1)+6\left(4\right)^(4-1)+6\left(4\right)^(5-1)=\\\\\\=6\cdot4^0+6\cdot4^1+6\cdot4^2+6\cdot4^3+6\cdot4^4=6+24+96+384+1536=\\\\=2,046`

or:

is geometric series:
`S_n=(a_1(1-q^n))/(1-q)`

`a_n=a_1\cdot q^(n-1)\quad\implies\quad\begin{cases}a_1=6\\q=4\\n=5\end{cases}`

so:

`\sum\limits_(k=1)^(5)6\left(4\right)^(k-1)=S_5=(6(1-4^5))/(1-4)=(6(1-1,024))/(-3)=-2(-1,023)=2,046`