# Write Out Each Sum

#### xyz_1965

See attachment. I need someone to get me started here.

#### romsek

MHF Helper
in general $\sum \limits_{k=0}^n \alpha^k = \dfrac{1-\alpha^{n+1}}{1-\alpha}$

For the second one

$\sum \limits_{k=3}^n (-1)^{k+1}2^k = \\ \sum \limits_{k=0}^{n-3} (-1)^{k+4}2^{k+3} = \\ \dfrac 1 2 \sum \limits_{k=0}^{n-3}(-2)^{k+4} = \\ 8\sum \limits_{k=0}^{n-3} (-2)^k = \\ 8 \dfrac{1-(-2)^{n-2}}{1-(-2)} = \\ \dfrac 8 3 \cdot \left(1 - (-2)^{n-2}\right)$

#### Debsta

MHF Helper
See attachment. I need someone to get me started here.

View attachment 39584
Although Romsek has given you an expression for each sum, are you simply meant to write this out as a sum eg 1+2+3+ ... +n sort of thing
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topsquark