- Prove that
![\[ \sum_{k=0}^n x^k = \frac{1-x^{n+1}}{1-x} \qquad \text{for } x \neq 1. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-3fe813fd6ddf48e7a2a676b41a67bdaf_l3.png "Rendered by QuickLaTeX.com")
Proof. We use properties of finite sums to compute,![\[ (1-x) \sum_{k=0}^n x^k = \sum_{k=0}^n (x^k - x^{k+1}) = - \sum_{k=0}^n (x^{k+1} -x^k) = -(x^{n+1} - 1) = 1-x^{n+1}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-ff24e463944fb514f7ebc8ed2c13e2ef_l3.png "Rendered by QuickLaTeX.com")
The second to last inequality follows from the telescoping property. But then solving for the sum we are interested in, we have
![\[ \sum_{k=0}^n x^k = \frac{1-x^{n+1}}{1-x}. \qquad \blacksquare \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-4980d746022060a702600f0f5eabf08e_l3.png "Rendered by QuickLaTeX.com")
- If
we have (using this, and the fact that 
for 
)
![\[ \sum_{k=0}^n x^k = \sum_{k=0}^n 1 = n+1. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-462635369cc03a173e7fced7d3019d20_l3.png "Rendered by QuickLaTeX.com")
The second to last inequality follows from the telescoping property. But then solving for the sum we are interested in, we have
we have (using this, and the fact that for )
[latextype]
for part b it states that \(\sum_{k=0}^{n} 1 = n + 1\) but in the referenced solution it shows that \( \sum_{k=0}^{n}1 = n \) so why is it different?
But in referenced solution it shows \( \sum_{k=1}^{n}1 = n \) Note that there is k = 1 and not k = 0 as is the case with this exercise. That’s why the solution is n + 1 and just n.