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Sum of the k-th powers of x for k = 0 to n

  1. Prove that

        \[ \sum_{k=0}^n x^k = \frac{1-x^{n+1}}{1-x} \qquad \text{for } x \neq 1. \]


    Proof. We use properties of finite sums to compute,

        \[ (1-x) \sum_{k=0}^n x^k = \sum_{k=0}^n (x^k - x^{k+1}) = - \sum_{k=0}^n (x^{k+1} -x^k) = -(x^{n+1} - 1) = 1-x^{n+1}. \]

    The second to last inequality follows from the telescoping property. But then solving for the sum we are interested in, we have

        \[ \sum_{k=0}^n x^k = \frac{1-x^{n+1}}{1-x}. \qquad \blacksquare \]

  2. If x = 1 we have (using this, and the fact that 1^k = 1 for k = 0, \ldots, n)

        \[ \sum_{k=0}^n x^k = \sum_{k=0}^n 1 = n+1. \]

One comment

  1. Anonymous says:

    [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?

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