Precalc
Section 9.3

Summation Notation

Sigma notation is the standard way to write series on Exam 4 β€” you'll see it in arithmetic and geometric sum problems, and it comes back in the Binomial Theorem. Getting comfortable reading and manipulating $\Sigma$ now means those problems won't slow you down.

⚑ Quick Summary

Summation notation is just a shorthand for "add up a bunch of things." The big Greek letter Ξ£\Sigma (sigma) means "sum." Instead of writing out 1+4+9+16+251 + 4 + 9 + 16 + 25, you can write βˆ‘i=15i2\sum_{i=1}^{5} i^2 β€” which says "plug in i=1,2,3,4,5i = 1, 2, 3, 4, 5 into i2i^2 and add up the results." It's like a compact instruction: the bottom tells you where to start, the top tells you where to stop, and the expression on the right tells you what to compute each time.

Sigma Notation
βˆ‘i=mnai=am+am+1+β‹―+an\sum_{i=m}^{n} a_i = a_m + a_{m+1} + \cdots + a_n

i is the index, m is the start, n is the end, aα΅’ is the expression

Sum of a Constant
βˆ‘i=1nc=cβ‹…n\sum_{i=1}^{n} c = c \cdot n

Adding the same number n times

πŸ“‹ Before You Start

Tap any item if you need a refresher:

Plain-English version

Summation notation is just a shorthand for "add up a bunch of things." The big Greek letter Ξ£\Sigma (sigma) means "sum." Instead of writing out 1+4+9+16+251 + 4 + 9 + 16 + 25, you can write βˆ‘i=15i2\sum_{i=1}^{5} i^2 β€” which says "plug in i=1,2,3,4,5i = 1, 2, 3, 4, 5 into i2i^2 and add up the results." It's like a compact instruction: the bottom tells you where to start, the top tells you where to stop, and the expression on the right tells you what to compute each time.

Sigma notation follows a few handy rules that let you break big sums into smaller, easier pieces. You can split a sum apart β€” βˆ‘(ai+bi)\sum(a_i + b_i) becomes βˆ‘ai+βˆ‘bi\sum a_i + \sum b_i. You can pull a constant multiplier out front β€” βˆ‘3i\sum 3i becomes 3β‹…βˆ‘i3 \cdot \sum i. These rules work just like distributing in regular algebra.

The real time-saver comes from shortcut formulas. Instead of adding 1+2+3+β‹―+1001 + 2 + 3 + \cdots + 100 one by one, you can use n(n+1)2\frac{n(n+1)}{2} and get 5050 instantly. There's a similar formula for the sum of squares. With these formulas plus the splitting rules, you can evaluate large sums without doing hundreds of additions.

One thing to watch: the shortcut formulas assume you start at i=1i = 1. If your sum starts at i=0i = 0 or i=2i = 2, you need to adjust β€” either handle the extra/missing terms separately or shift the formula. When the upper limit is small (like 4 or 5), it's often faster to just plug in each value and add by hand rather than use the formulas.

Key Formulas

Sigma Notation
βˆ‘i=mnai=am+am+1+β‹―+an\sum_{i=m}^{n} a_i = a_m + a_{m+1} + \cdots + a_n

i is the index, m is the start, n is the end, aα΅’ is the expression

Sum of a Constant
βˆ‘i=1nc=cβ‹…n\sum_{i=1}^{n} c = c \cdot n

Adding the same number n times

Sum of First n Positive Integers
βˆ‘i=1ni=n(n+1)2\sum_{i=1}^{n} i = \frac{n(n+1)}{2}

The classic Gauss formula: 1 + 2 + 3 + β‹― + n

Sum of First n Squares
βˆ‘i=1ni2=n(n+1)(2n+1)6\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}
Linearity of Summation
βˆ‘i=1n(ai+bi)=βˆ‘i=1nai+βˆ‘i=1nbi,βˆ‘i=1ncβ‹…ai=cβ‹…βˆ‘i=1nai\sum_{i=1}^{n}(a_i + b_i) = \sum_{i=1}^{n} a_i + \sum_{i=1}^{n} b_i, \qquad \sum_{i=1}^{n} c \cdot a_i = c \cdot \sum_{i=1}^{n} a_i

Split sums apart and pull constants out β€” just like distributing

Key Takeaways

  • βœ“Three formulas to memorize: βˆ‘c=cn\sum c = cn, βˆ‘i=n(n+1)2\sum i = \frac{n(n+1)}{2}, βˆ‘i2=n(n+1)(2n+1)6\sum i^2 = \frac{n(n+1)(2n+1)}{6}
  • βœ“Linearity: split sums apart and pull constants out β€” βˆ‘(ai+bi)=βˆ‘ai+βˆ‘bi\sum(a_i + b_i) = \sum a_i + \sum b_i
  • βœ“Always check the starting index β€” formulas assume i=1i = 1; adjust if the sum starts at 00 or 22
  • βœ“For small upper limits (4 or 5), just expand and add β€” it's faster than using formulas

⚠️ Common Mistakes

  • βœ—Forgetting to check the starting index β€” most closed-form formulas assume i=1i = 1, so if your sum starts at i=0i = 0 or i=2i = 2, you need to adjust or compute the extra/missing terms separately
  • βœ—Trying to pull out a variable as if it were a constant β€” βˆ‘3i2\sum 3i^2 lets you pull out the 33, but βˆ‘iβ‹…i2\sum i \cdot i^2 does NOT let you pull out the ii because ii changes with each term
  • βœ—Mixing up the closed-form formulas β€” βˆ‘i=n(n+1)2\sum i = \frac{n(n+1)}{2} vs. βˆ‘i2=n(n+1)(2n+1)6\sum i^2 = \frac{n(n+1)(2n+1)}{6}; double-check which one you need before plugging in
  • βœ—Forgetting that βˆ‘i=1nc=cn\sum_{i=1}^{n} c = cn, not just cc β€” a constant summed nn times gives cncn