Adding up the terms of a sequence

Series and partial sums S(n), recovering the terms from the partial sums with T(n) = S(n) − S(n−1), and reading and writing sums in sigma notation.

Partial sums

Adding up the terms of a sequence gives a series. The sum of the first $n$ terms is the partial sum $S_n = T_1 + T_2 + \cdots + T_n$.

For $1, 3, 5, 7, 9, \ldots$ the partial sums are $1, 4, 9, 16, 25, \ldots$ (the square numbers).

Example 1●●

The partial sums of a sequence are $S_n = 3n^2 + n$. Find $T_1$, $T_{10}$ and a formula for $T_n$.

$T_1 = S_1 = 4$.

$T_{10} = S_{10} - S_9 = 310 - 252 = 58$.

$T_n = (3n^2 + n) - \big(3(n - 1)^2 + (n - 1)\big) = 6n - 2$. It also gives $T_1 = 4$.

Sigma notation

The Greek capital sigma $\Sigma$ means "add up". The expression $$\sum_{k=1}^{5} k^2$$ means: let $k$ run through $1, 2, 3, 4, 5$, work out $k^2$ each time, and add the results. So it equals $1 + 4 + 9 + 16 + 25 = 55$.

Example 2●

Evaluate $\displaystyle\sum_{k=2}^{5} (3k - 1)$.

$5 + 8 + 11 + 14 = 38$

Example 3: writing a sum in sigma notation●●

Write $3 + 6 + 12 + \cdots + 768$ in sigma notation.

The terms are $3 \times 2^{k-1}$, and $768 = 3 \times 256 = 3 \times 2^8$, so the last value of $k$ is $9$.

$\displaystyle\sum_{k=1}^{9} 3 \times 2^{k-1}$

Flashcards

$S_n$

$T_1 + T_2 + \cdots + T_n$

$T_n$ from $S_n$

$T_n = S_n - S_{n-1}$, and $T_1 = S_1$.

Terms in $displaystylesum_{k=m}^{n}$

$n - m + 1$

Practice

If $S_n = 3n^2 + n$, find $T_{10}$.
Evaluate $\displaystyle\sum_{k=1}^{5} k^2$.
Evaluate $\displaystyle\sum_{k=2}^{5} (3k - 1)$.
How many terms are in $\displaystyle\sum_{k=3}^{10} k^3$?
If $S_n = 3n^2 + n$, then $T_n = $
Which sum equals $2 + 4 + 6 + \cdots + 40$?
Evaluate $\displaystyle\sum_{k=1}^{4} (-1)^k k$.
The partial sums of $1, 3, 5, 7, \ldots$ are square numbers. Find $S_5$.
If $S_n = n^2 + 3n$, find $T_1$.