
\newpage

\textbox{
Here are some standard summation formulas that you should know.

\begin{align*}
\sum_{i=0}^n i & = \frac{n(n+1)}{2}
	& \text{arithmetic sum} \\
\sum_{i=0}^n i^2 & = \frac{n(n+1)(2n+1)}{6}
	& \text{sum of squares} \\
\sum_{i=0}^n z^i & = \frac{1 - z^{n+1}}{1 - z} \qquad (z \neq 1)
	& \text{geometric sum} \\
\sum_{i=0}^{\infty} z^i & = \frac{1}{1 - z} \qquad (|z| < 1)
	& \text{infinite geometric sum}
\end{align*}
%
Differentiating the geometric sum formula and multiplying both sides
by $z$ gives:
%
\[
\sum_{i=0}^n i z^i
    = \frac{z - (n+1) z^{n+1} + n z^{n+2}}{(1-z)^2} \qquad (z \neq 1)
\]
}

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\newpage

\begin{problem}
Find closed-form expressions equal to the following sums.  You do not
need to simplify your answers.

\bparts

\ppart
\[
\sum_{k=0}^n k^2 + k + 1
\]

\solution[\vspace{2in}]{We break the sum into three separate sums so
that the standard formulas apply.

\begin{align*}
\sum_{k=0}^n k^2 + k + 1
    & = \left(\sum_{k=0}^n k^2 \right) +
        \left(\sum_{k=0}^n k \right) +
        \left(\sum_{k=0}^n 1 \right) \\
    & = \frac{n (n+1) (2n + 1)}{6} + \frac{n (n+1)}{2} + n
\end{align*}
}

\ppart
\[
1 - \frac{1}{10} + \frac{1}{100} - \frac{1}{1000} + \frac{1}{10000} - \ldots
\]

\solution[\vspace{2in}]{
This is an infinite geometric sum with ratio $-1/10$:
%
\begin{align*}
\sum_{k=0}^{\infty} \left(- \frac{1}{10} \right)^k
    & = \frac{1}{1 - (- 1/10)} \\
    & = \frac{10}{11}
\end{align*}
}

\ppart
\[
\sum_{k=0}^n (3 k + 1) 2^k
\]

\solution{
Again, we break up the sum so that standard formulas apply:
%
\begin{align*}
\sum_{k=0}^n (3 k + 1) 2^k
    & = 3 \cdot \left( \sum_{k=0}^n k 2^k \right) + \sum_{k=0}^n 2^k \\
    & = 3 \cdot \left(\frac{z - (n+1) z^{n+1} + n z^{n+2}}{(1-z)^2}\right)
        + \frac{1 - 2^{n+1}}{1 - 2}
\end{align*}
}

\eparts

\end{problem}

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