## anonymous one year ago Evaluate the series: Enter a numerical answer only. ∑8k = 4(2k−1)

1. anonymous

Something is wrong in the question, perhaps the series is dual to 4(2n-1) such that n is the last term in the series

2. anonymous

Equal**

3. anonymous

$\frac{ 8k \left( 8k+1 \right) }{ 2 }=4(2k-1)$

4. anonymous

|dw:1435181961855:dw|

5. anonymous

K is varying in the series, and so it has to be in the other side which is impossible, it could be an equation with n in the right hand side

6. anonymous

$\sum_{k=4}^{8}\left( 2k-1 \right)=7+9+11+13+15=\frac{ 5 }{ 2 }\left( 7+15 \right)$