Summation Notation Makes Repeated Addition Precise

Long additions hide their structure. Summation notation—or sigma notation—states what changes, what stays fixed, where counting begins, and where it ends.

Learning objectives

You will learn to expand and compress sums, distinguish an index from a term count, use linearity, shift indices, and apply standard finite-sum formulas.

Read the notation from the inside out

For

$$\sum_{k=1}^{4}(2k+1),$$

substitute $k=1,2,3,4$ into the entire summand:

$$(3)+(5)+(7)+(9)=24.$$

There are $n-m+1$ terms in an inclusive sum from $m$ through $n$. Thus $\sum_{i=3}^{7}a_i$ contains five terms, not four or seven.

Linearity of summation

Constants distribute through finite sums:

$$\sum_{i=m}^{n}(ca_i+db_i) =c\sum_{i=m}^{n}a_i+d\sum_{i=m}^{n}b_i.$$

Also,

$$\sum_{i=m}^{n}c=(n-m+1)c.$$

Standard formulas and why they matter

Three common identities are

$$\sum_{i=1}^{n}1=n, \qquad \sum_{i=1}^{n}i=\frac{n(n+1)}{2}, \qquad \sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}.$$

The first counts terms. The second can be understood by pairing the first and last terms: each pair sums to $n+1$. These formulas turn structured additions into algebraic expressions and later make Riemann sums computable.

Index shifts preserve the terms

Consider

$$\sum_{i=1}^{n}a_{i+1}=a_2+a_3+\cdots+a_{n+1}.$$

Let $j=i+1$. Then $i=1$ corresponds to $j=2$, and $i=n$ corresponds to $j=n+1$:

$$\sum_{i=1}^{n}a_{i+1}=\sum_{j=2}^{n+1}a_j.$$

Change the index, bounds, and summand together. Expanding the first few terms is the safest check.

Nested sums

In

$$\sum_{i=1}^{2}\sum_{j=1}^{3}(i+j),$$

the inner sum completes for each fixed $i$. For $i=1$, it gives $2+3+4=9$; for $i=2$, it gives $3+4+5=12$. The total is $21$.

Check your understanding

1. Expand $\sum_{r=0}^{3}2^r$.
2. How many terms are in $\sum_{k=4}^{12}b_k$?
3. Rewrite $\sum_{i=0}^{n-1}a_{i+1}$ using index $j$ beginning at $1$.

Show the reasoning

1. $2^0+2^1+2^2+2^3=1+2+4+8=15$.
2. $12-4+1=9$ terms.
3. Set $j=i+1$. The bounds become $j=1$ through $j=n$, so the sum is $\sum_{j=1}^{n}a_j$.