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Arithmetic sequences

Consecutive terms differ by a constant $d$ :

a_n = a_1 + (n-1)d, \qquad S_n = \dfrac{n}{2}\big(2a_1 + (n-1)d\big) = \dfrac{n}{2}(a_1 + a_n)

Geometric sequences

Consecutive terms share a constant ratio $r$ :

a_n = a_1 r^{\,n-1}, \qquad S_n = \dfrac{a_1(1-r^n)}{1-r} \ (r \neq 1), \qquad S_\infty = \dfrac{a_1}{1-r}\ \ (|r|<1)

The infinite-sum formula only applies when $|r| < 1$ — otherwise the terms don't shrink and the sum diverges.

Telescoping sums

A sum “telescopes” when each term can be rewritten as a difference of two pieces such that consecutive terms cancel, leaving only the first and last piece standing. The classic setup:

\sum_{k=1}^{n} \left(\dfrac{1}{k} - \dfrac{1}{k+1}\right) = \dfrac{1}{1} - \dfrac{1}{n+1}

A useful trick for spotting this: if a term looks like $\tfrac{1}{k(k+1)}$ , use partial fractions ( $\tfrac{1}{k(k+1)} = \tfrac1k - \tfrac1{k+1}$ ) to reveal the telescoping structure.

Worked example

Problem: Evaluate $\displaystyle\sum_{k=1}^{99} \dfrac{1}{k(k+1)}$ .

Solution: Split each term with partial fractions: $\dfrac{1}{k(k+1)} = \dfrac1k - \dfrac{1}{k+1}$ . The sum becomes

\left(1-\tfrac12\right)+\left(\tfrac12-\tfrac13\right)+\cdots+\left(\tfrac{1}{99}-\tfrac{1}{100}\right)

Every middle term cancels with its neighbor, leaving just $1 - \tfrac{1}{100} = \tfrac{99}{100}$ .

Practice problems

1.An arithmetic sequence has $a_1=5$ and $d=3$ . Find $a_{20}$ .

2.Find the sum of the first 10 terms of the geometric sequence $3, 6, 12, 24, \ldots$

3.Find the sum of the infinite geometric series $4 + 2 + 1 + \tfrac12 + \cdots$

4.Evaluate $\displaystyle\sum_{k=1}^{50}\left(\sqrt{k+1}-\sqrt{k}\right)$ .

Sequences & Recursions

Arithmetic sequences

Geometric sequences

Telescoping sums

Practice problems