Number Patterns

Problem 1. Consider the following equations.

$\begin{aligned}T_1 &= 1 = \textstyle \frac 12 \times 1 \times 2, \\ T_2 &= 1 + 2 = \textstyle \frac 12 \times 2 \times 3, \\ T_3 &= 1+2+3 = \textstyle \frac 12 \times 3 \times 4.\end{aligned}$

Write down equations for $T_4, T_5$ . Hence, write down an equation for $T_n$ in terms of $n$ and prove that this equation is correct.

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Solution. We check that

$\begin{aligned} T_4 &= 1 + 2 + 3 + 4 = \textstyle \frac 12 \times 4 \times 5 = 10, \\ T_5 &= 1 + 2 + 3 + 4 + 5= \textstyle \frac 12 \times 5 \times 6 = 15.\end{aligned}$

Hence, we conjecture $T_n = \frac 12 \times n \times (n+1)$ , and prove it as follows:

$\begin{aligned} T_n &= 1 + 2 + 3 + \cdots + (n-2) + (n-1) + n \\ T_n &= n + (n-1) + (n-2) + \cdots + 3 + 2 + 1 \\ T_n +T_n &= \underbrace{ (n+1) + (n+1) + \cdots + (n+1) + (n+1) }_n \\ 2T_n &= n \times (n+1) \\ T_n &= \textstyle \frac 12 \times n \times (n+1).\end{aligned}$

Problem 2. Let $a, d$ be real numbers. For each $k$ , define

$u_k = a + (k-1) d.$

Given a positive integer $n > 1$ , use Problem 1 to evaluate

$S_n := u_1 + u_2 + \cdots + u_n$

in terms of $n$ .

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Solution. By expanding the terms,

$\begin{aligned} S_n &= a + (a+d) + \cdots + (a+(n-1)d) \\ &= (\underbrace{a + a + \cdots + a}_n) + ( 1 + 2 + \cdots + (n-1) )d \\ &= na + T_{n-1} d \\ &= na + \textstyle \frac 12 (n-1)nd \\ &= \textstyle \frac 12n(2a + (n-1)d) \\ &= \textstyle \frac 12 n(u_1 + u_n). \end{aligned}$

Remark 1. The sequence $\{ u_k \}$ in Problem 2 is called an arithmetic progression with first term $a$ and common difference $d$ . The sum of the first $n$ terms of such a progression is given by $S_n$ .

Problem 3. Fix a real number $r \neq 1$ . Consider the following equations.

$\begin{aligned}T_1 &= 1 + r = \frac{1 - r^2}{1 - r}, \\ T_2 &= 1+r+r^2 = \frac{1 - r^3}{1 - r} ,\\ T_3 &= 1+r+r^2+r^3 = \frac{1 - r^4}{1 - r}.\end{aligned}$

Write down an equation for $T_n$ in terms of $n$ and prove that this equation is correct.

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Solution. By following the pattern, we conjecture that

$\displaystyle T_n = 1 + r + r^2 + \cdots + r^n = \frac{ 1 - r^{n+1} }{1-r}.$

We prove this result as follows:

$\begin{aligned} T_n &= 1 + r + r^2 + \cdots + r^n \\ rT_n &= \phantom{1+.\!} r + r^2 + \cdots + r^n + r^{n+1} \\ T_n - rT_n &= 1 - r^{n+1} \\ (1-r)T_{n} &= 1 - r^{n+1} \\ T_{n} &= \frac{ 1 - r^{n+1} }{1 - r}. \end{aligned}$

Problem 4. Let $a \neq 0$ and $r \neq 1$ be real numbers. For each $k$ , define

$v_k = a r^{k-1}$

Given a positive integer $n > 1$ , use Problem 2 to evaluate

$W_n := v_1 + v_2 + \cdots + v_n$

in terms of $n$ .

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Solution. By expanding the terms,

$\begin{aligned} W_n &= u_1 + u_2 + \cdots + u_n \\ &= a + ar + \cdots + ar^{n-1} \\ &= a(1 + r + \cdots + r^{n-1}) \\ &= a \cdot T_{n-1} \\ &= \frac{a(1 - r^n)}{1-r}. \end{aligned}$

Remark 1. The sequence $\{ u_k \}$ in Problem 4 is called a geometric progression with first term $a$ and common ratio $r$ . The sum of the first $n$ terms of such a progression is given by $T_n$ .

Problem 5. Show that for any $n > 1$ , $u_n - u_{n-1}$ is constant and $v_n/v_{n-1}$ is constant.

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Solution. By Problems 2 and 4,

$\begin{aligned} u_n - u_{n-1} &= a + (n-1) d - (a + (n-2)d) \\ &= ((n-1) - (n-2))d \\ &= d, \\ \frac{ v_n }{ v_{n-1}} &= \frac{ar^{n-1}}{ar^{n-2}} = r. \end{aligned}$

Problem 6. Let $\{x_n\}$ be a sequence of nonzero numbers such that for any $n > 1$ , $x_n - x_{n-1}$ is constant and $x_n / x_{n-1}$ is constant. Show that $x_i = x_j$ for any $i, j$ .

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Solution. Suppose there exists real numbers $d, r$ such that for any $n > 1$ ,

$\displaystyle x_n - x_{n-1} = d,\quad \frac{x_n}{x_{n-1}} = r.$

Since $x_n = r \cdot x_{n-1}$ , we have

$\begin{aligned} (r-1) \cdot x_{n-1} &= r \cdot x_{n-1} - x_{n-1} \\ &= d. \end{aligned}$

Using the same logic but replacing $n$ with $n+1$ ,

$\displaystyle (r-1) \cdot x_{n} = d.$

Subtracting both equations,

$\begin{aligned} (r-1) \cdot d &= (r-1) \cdot (x_n - x_{n-1}) \\ &= d-d \\ &= 0. \end{aligned}$

Therefore, either $r = 1$ or $d = 0$ . In both cases, $x_n = x_{n-1}$ . Since this reasoning works for any $n > 1$ , we have $x_n = x_1$ for any $n > 1$ . In particular, $x_i = x_j$ for any $i, j$ .