Summation notation is a way of writing a series such as

\[1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\cdots+\frac{1}{n^2}+\cdots\]

in a shorter form. The shorter form captures the fact that it is a *sum* by using
the greek letter capital sigma \(\Sigma\), indicates the index, in this case the letter
\(n\), the starting and ending values of the index \(n=1\) and infinity \(\infty\) (to
indicate that the index increases one unit at a time without bound) and finally, the
formula for the general term \(a_n=\frac{1}{n^2}\).

\[1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\cdots+\frac{1}{n^2}+\cdots=\sum_{n=1}^\infty\frac{1}{n^2}\]

Summation notation is analogous to sequence notation \(\{a_n\}\) with the exception that in sequence notation the sequence usually starts with \(n=1\) and procedes indefinitely, so it is unnecessary to specify that \(n\) goes from \(1\) to infinity.

Rewrite the *finite* series (\(n\) does not go to infinity) using summation notation:
\(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+\cdots+\dfrac{50}{51}\)

Calculate the value of the sum:\(\quad\displaystyle\sum_{p=1}^4(p-1)^2\)

Calculate the value of the sum:\(\quad\displaystyle\sum_{k=3}^7(k^2-2)\)

Write the following as a single sum:\(\quad\displaystyle\sum_{p=0}^{3}p^2+\sum_{p=4}^{6}p^2\)

Write the following as a single sum:\(\quad\displaystyle\sum_{p=0}^{3}p^2+\sum_{q=4}^{6}q^2\)

Suppose \(f(x)=1-x+x^2-x^3\). We could use summation notation and write

\[f(x=\sum_{n=0}^3(-1)^nx^n\]

Take special notice of how the factor \((-1)^n\) causes the sum to *alternate in sign*.

Write the function in summation notation: \(f(x)=1-\frac{1}{2}x+\frac{1}{3}x^2-\cdots+\frac{1}{101}x^{100}\)