Sequences & Series

Find the first four terms of $a_n = n^2 - 1$.

The sequence is $0,\ 3,\ 8,\ 15,\ \dots$

Find the first four terms of $a_1 = 5,\ a_n = a_{n-1} + 4$.

The sequence is $5,\ 9,\ 13,\ 17,\ \dots$

For $5,\ 8,\ 11,\ 14,\ \dots$ find $a_{10}$.

Find $S_{10}$ for $5,\ 8,\ 11,\ 14,\ \dots$ From the previous section, $a_{10} = 32$.

Find $2 + 4 + 6 + \cdots + 100$ (the first $50$ positive even numbers). Here $a_1 = 2$, $a_{50} = 100$, $n = 50$.

For $3,\ 6,\ 12,\ 24,\ \dots$ find $a_7$.

Find $S_6$ for $3,\ 6,\ 12,\ 24,\ \dots$ where $a_1 = 3$ and $r = 2$.

Find $8 + 4 + 2 + 1 + \cdots$ Here $a_1 = 8$ and $r = \frac{4}{8} = \frac{1}{2}$. Since $|r| < 1$, the sum exists.

Write $0.\overline{7} = 0.7 + 0.07 + 0.007 + \cdots$ as a fraction. This is geometric with $a_1 = 0.7$ and $r = 0.1$.

Evaluate $\displaystyle\sum_{n=1}^{4}(2n+1)$.

Write $3 + 6 + 9 + 12 + 15$ using sigma notation. Each term is $3n$ for $n = 1$ to $5$:

A machine worth $50,000 loses $20\%$ of its value each year, so it keeps $80\%$: $r = 0.8$. Find its value after $5$ years, i.e. the term $a_n = 50000\,(0.8)^{n}$ at $n = 5$.

A ball dropped from $10$ m rebounds to $60\%$ of its previous height. Find the total vertical distance it travels before resting. It falls $10$ m, then each rebound goes up and back down. The rebound heights $6, 3.6, 2.16, \dots$ are geometric with $a_1 = 6$, $r = 0.6$.

The factor of $2$ counts each rebound up and down; the initial $10$ m drop is counted once.

You start at $35,000 and receive a $1,500 raise each year. Total earned over $10$ years? Arithmetic with $a_1 = 35000$, $d = 1500$.

Evaluate $\displaystyle\sum_{k=1}^{n}\frac{1}{k(k+1)}$, then take $n \to \infty$. Split each term with partial fractions:

As $n \to \infty$, $\dfrac{1}{n+1} \to 0$, so the infinite sum equals $\boxed{1}$.

Terms like $k\,r^k$ mix an arithmetic factor ($k$) with a geometric one ($r^k$). Evaluate $\displaystyle S = \sum_{k=1}^{\infty} k\left(\tfrac12\right)^{k}$ using the shift-and-subtract trick. First check convergence: here $r = \tfrac12$, and since $|r| < 1$ the sum exists. Write the finite sum $S_n = \sum_{k=1}^{n} k r^k$ and subtract $rS_n$:

With $|r| < 1$, both $r^n \to 0$ and $nr^{n+1} \to 0$ as $n \to \infty$, leaving the tidy formula

Now substitute $r = \tfrac12$:

So $\tfrac12 + \tfrac24 + \tfrac38 + \tfrac{4}{16} + \cdots = \boxed{2}$.

Turn the recursive rule $a_1 = 3,\ a_n = 2a_{n-1} + 1$ into an explicit formula. First find the fixed point $L$ that the constant term wants to settle at, by solving $L = 2L + 1$, which gives $L = -1$. Add $1$ to both sides of the recurrence and the “$+1$” disappears:

So the shifted sequence $b_n = a_n + 1$ is geometric with ratio $2$ and first term $b_1 = a_1 + 1 = 4$. Therefore

Check: $a_1 = 2^2 - 1 = 3$, $a_2 = 2^3 - 1 = 7$, $a_3 = 2^4 - 1 = 15$. Matches $3, 7, 15, 31, \dots$

Claim. For every integer $n \ge 1$, $\displaystyle\sum_{k=1}^{n}(2k-1) = n^2$ (the sum of the first $n$ odd numbers).

Base case ($n = 1$). The left side is $2(1) - 1 = 1$; the right side is $1^2 = 1$. They agree, so $P(1)$ holds.

Inductive step. Assume the formula holds for some $n = k$; that is, assume

We must show it holds for $n = k+1$, i.e. that the sum equals $(k+1)^2$. Start from the $(k+1)$-term sum and peel off the last term:

This is exactly $P(k+1)$. By the principle of mathematical induction, the formula holds for all $n \ge 1$. $\blacksquare$

An endowment pays $100 at the end of every year forever, and money is discounted at $i = 5\%$ per year. Its present value is the sum of all payments discounted back to today:

This is geometric with $a_1 = \dfrac{100}{1.05}$ and $r = \dfrac{1}{1.05}$. Since $i > 0$ we have $r < 1$, so the sum converges:

A perpetual stream of $100 per year is worth just $2000 today — a clean shortcut: $\text{PV} = \dfrac{\text{payment}}{i}$.