Number pyramids

Stack a row of numbers, then build upward: each cell above equals the sum of the two cells directly touching it below. Fill in a bottom row and the entire pyramid — every cell all the way to the single number at the top — is fixed. That is a fun trick to show off, but the real question a Class 8 student should ask is why it always works, and how the top number depends on the bottom ones. This is where algebra earns its keep: instead of adding numbers one pyramid at a time, we assign letters to the unknown bottom entries, climb the pyramid symbolically using the distributive law, and end up with a single formula for the top cell. That formula lets us predict, prove, and — most usefully for exams — solve for a missing number when the top and some other entries are already known.

The core insight

In a number pyramid, algebra turns "each cell is the sum of the two below it" into one combined formula for the top cell in terms of the bottom-row letters, which can then be solved like any linear equation.

The core idea

The building rule. Every cell in a number pyramid, except the bottom row, equals the sum of the two cells directly beneath it — never the diagonal cells, and never cells further away.

Turning the pyramid into algebra. Assign a letter to each bottom-row entry, then build upward one row at a time, simplifying with the distributive law as you go. For a 3-row pyramid with bottom row $a$ , $b$ , $c$ , the middle row is $a+b$ and $b+c$ , and the top cell is their sum:

\text{top} = (a+b) + (b+c) = a + 2b + c

where: $a$ , $b$ , $c$ are the three bottom-row numbers (known or unknown), and the middle coefficient $2$ appears because $b$ is used in building both middle-row cells.

Bigger pyramids follow a pattern. Add a row and the coefficients grow according to Pascal's-triangle-style weights: $1,1$ for a 2-number base, $1,2,1$ for 3 numbers, $1,3,3,1$ for 4 numbers, and so on. So a 4-row pyramid with base $a,b,c,d$ has top cell $a+3b+3c+d$ . A curious student can verify this by building the pyramid row by row rather than memorising the coefficients — memorising invites the common trap of guessing the wrong weights.

From formula to equation. Whenever the top cell (or another cell) is given and a bottom-row number is unknown, the algebraic expression for that cell becomes a linear equation, which is solved the usual way: combine like terms, then undo addition/subtraction and multiplication/division in reverse order.

Figure

A 3-row number pyramid showing bottom cells a b c the middle cells a+b and b+c formed from the two cells below and the single top cell a+2b+c formed from the middle row

The top-cell formula

Derivation AThe top cell of a 3-row number pyramid

The goal is to derive a single algebraic formula for the top cell of a 3-row number pyramid in terms of its three bottom-row entries, using nothing but the pyramid's own building rule and the distributive law, so that the formula can later be set equal to a known top value and solved as a linear equation for a missing bottom entry.

Step 1. Name the bottom row. Let the three bottom-row entries be $a$ , $b$ , $c$ , read left to right. These may be given numbers, or one of them may be an unknown such as $x$ .

Step 2. Build the middle row using the pyramid rule "each cell equals the sum of the two cells directly below it." The left middle cell sits above $a$ and $b$ , so it equals $a+b$ . The right middle cell sits above $b$ and $c$ , so it equals $b+c$ .

Step 3. Build the top cell the same way: it sits above the two middle cells, so it equals their sum, $(a+b)+(b+c)$ .

Step 4. Simplify using the distributive law — remove the brackets and combine like terms: $(a+b)+(b+c) = a + b + b + c = a + 2b + c$ . The two copies of $b$ combine into $2b$ because $b$ was used in building both middle cells, while $a$ and $c$ each appear only once, in only one middle cell.

Step 5. State the result: for any 3-row pyramid with bottom row $a$ , $b$ , $c$ , \text{top} = a + 2b + c This holds for any numbers in the bottom row, so it can be used both to compute an unknown top cell from a known bottom row, and — by setting the expression equal to a given top value — to solve for an unknown bottom entry.

Worked example

Problem. In a 3-row pyramid, the bottom row is $5$ , $x$ , $9$ . The top cell is known to be $30$ . Find $x$ .

Step 1. Apply the formula top $=a+2b+c$ with $a=5$ , $b=x$ , $c=9$ : 5 + 2x + 9 = 30

Step 2. Combine the constant terms on the left: 2x + 14 = 30

Step 3. Subtract $14$ from both sides: 2x = 16

Step 4. Divide both sides by $2$ : x = 8

Step 5. Check by rebuilding the pyramid with bottom row $5, 8, 9$ . Middle row: $5+8=13$ and $8+9=17$ . Top: $13+17=30$ , which matches the given top cell, confirming $x = 8$ .

Competitive stretch

A trickier question asks students to test a claimed pattern rather than just solve for $x$ . Take a bottom row of three consecutive numbers, $a$ , $a+1$ , $a+2$ . Building up: middle row is $a+(a+1)=2a+1$ and $(a+1)+(a+2)=2a+3$ ; top cell is $(2a+1)+(2a+3)=4a+4$ . Since the middle bottom number is $a+1$ , this simplifies to $4a+4 = 4(a+1)$ — so the top cell always equals 4 times the middle bottom number whenever the base is three consecutive numbers. Checking with $3,4,5$ : middle row $7,9$ ; top $16 = 4\times4$ . Checking with $10,11,12$ : middle row $21,23$ ; top $44=4\times11$ . Both confirm the pattern algebraically, not just by luck on one example.

Common traps

Adding diagonal cells instead of the two directly below. The rule only ever refers to the two cells immediately underneath, never diagonal or distant ones.
Guessing the coefficients for a bigger pyramid instead of building it. A 4-row pyramid gives top $=a+3b+3c+d$ , not $a+2b+2c+d$ — get this from building row by row, not from memory.
Trusting one numerical check as proof. A pattern that looks right for a single example can still be mis-stated; always simplify the general algebraic expression before declaring a pattern true.
Sign or combining-like-terms slips, such as writing $(2a+1)+(2a+3)$ as $4a+3$ instead of the correct $4a+4$ .

Ask the tutor: Stuck on a pyramid with more rows or two unknown bottom cells — ask the AI tutor to build it up with you step by step.

Why does the middle number get a coefficient of 2 in a 3-row pyramid but a and c do not

Because of how many times each bottom number is actually used while climbing the pyramid. The middle number $b$ feeds into both middle-row cells — the left one ( $a+b$ ) and the right one ( $b+c$ ) — so when those two cells are added to make the top, $b$ appears twice and combines into $2b$ . The end numbers $a$ and $c$ each feed into only one middle-row cell, so they appear just once in the top-cell expression. Building the pyramid row by row, rather than memorising the formula, makes this visible.

Do I have to memorise the coefficient patterns for 4-row and 5-row pyramids

No — and memorising them without understanding is risky, because it is easy to misremember which row goes with which set of coefficients. It is safer and just as fast to build the pyramid one row at a time using the basic rule, sum of the two cells below, and simplify as you go. The Pascal's-triangle-style pattern ( $1,1$ then $1,2,1$ then $1,3,3,1$ ...) is a nice thing to notice once you have built a few pyramids, but the reliable method for any exam question is always to build up from the bottom row.

What if two different bottom-row letters are unknown, not just one

The same method still works, but you will need two equations to pin down two unknowns — usually from two given cells (for example, the top cell and one middle cell), or from an extra condition stated in the problem, such as the bottom-row numbers being consecutive or equal. Write the algebraic expression for each given cell in terms of the unknowns, then solve the resulting pair of equations together, the same way you would solve any pair of linear equations in two variables.