Pascal's Triangle and the Binomial Pattern

A Pattern Worth Noticing

You already know these identities:

$$(a+b)^0 = 1, \quad (a+b)^1 = a+b, \quad (a+b)^2 = a^2 + 2ab + b^2$$

$$(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3, \quad (a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$$

Computing $(a+b)^4$ by multiplying $(a+b)^3$ by $(a+b)$ works, but it hides a structure worth spelling out in three observations:

Key Point (the three observations): In the expansion of $(a+b)^n$: (1) the number of terms is one more than the index — $n + 1$ terms; (2) powers of $a$ decrease by 1 term by term while powers of $b$ increase by 1; (3) in every term the indices of $a$ and $b$ add to $n$.

So the only thing left to know is the coefficients — and they have a life of their own.

Pascal's Triangle

Arrange the coefficients by index:

Every row begins and ends with 1, and each interior entry is the sum of the two entries above it — the 2 in row 2 comes from $1 + 1$; the two 3's in row 3 from $1+2$ and $2+1$. This array is Pascal's triangle, known to the Indian scholar Pingla as Meru Prastara centuries before Pascal.

To expand a binomial, read off its row. The row for index 5 is $1, 5, 10, 10, 5, 1$, so

$$(2x + 3y)^5 = (2x)^5 + 5(2x)^4(3y) + 10(2x)^3(3y)^2 + 10(2x)^2(3y)^3 + 5(2x)(3y)^4 + (3y)^5$$

$$= 32x^5 + 240x^4y + 720x^3y^2 + 1080x^2y^3 + 810xy^4 + 243y^5$$

[Board Tip] The whole quantity $2x$ carries each power — $(2x)^3 = 8x^3$, not $2x^3$. Forgetting the coefficient's power is this section's classic lost mark.

From Triangle Rows to Combinations

Expanding $(2x+3y)^{12}$ by triangle would need all twelve rows written out — workable, but slow, and hopeless for index 50. The escape: the entries of Pascal's triangle are the combinations of Chapter 6.

Row $n$ reads $^nC_0, {^nC_1}, \ldots, {^nC_n}$ — so any row is available on demand. For index 7: $^7C_0, \ldots, {^7C_7} = 1, 7, 21, 35, 35, 21, 7, 1$, giving

$$(a+b)^7 = {^7C_0}a^7 + {^7C_1}a^6b + {^7C_2}a^5b^2 + \cdots + {^7C_7}b^7$$

Key Point: The triangle's addition rule is exactly the Pascal identity $^nC_r + {^nC_{r-1}} = {^{n+1}C_r}$ proved in Chapter 6 — the two chapters are one story.

[JEE Tip] Memorise rows up to index 6 ($1,6,15,20,15,6,1$) — most exam expansions live there, and recognising $1, 5, 10, 10, 5, 1$ on sight saves a minute per question.

Solved Examples

Example 1: Reading the observations

Without expanding, state for $(p + q)^9$: (i) the number of terms (ii) the power of $p$ in the 4th term (iii) the sum of indices in every term.

Solution:

Step 1 — (i) Count the terms. The expansion of $(a+b)^n$ has $n + 1$ terms: $9 + 1 = 10$.

Step 2 — (ii) Track the falling power. Powers of $p$ run $9, 8, 7, 6, \ldots$ from the first term — the 4th term carries $p^6$.

Step 3 — (iii) Index sum. In every term the indices of $p$ and $q$ add to the index: always $9$.

Takeaway: The three observations answer structural questions with zero computation.

Example 2: Building a row

Given the row for index 6 is $1, 6, 15, 20, 15, 6, 1$, write the row for index 7.

Solution:

Step 1 — Apply the addition rule. Each new interior entry is the sum of the two entries above: $1+6, \ 6+15, \ 15+20, \ 20+15, \ 15+6, \ 6+1$.

Step 2 — Pad with the boundary 1's. Row 7: $1, 7, 21, 35, 35, 21, 7, 1$.

Step 3 — Check. These are $^7C_0, \ldots, {^7C_7}$, and the row sums to $2^7 = 128$: $1+7+21+35+35+21+7+1 = 128$ ✓.

Takeaway: Adjacent sums build the next row — and $^nC_r$ values confirm it.

Example 3: A fifth-power expansion

Expand $(2x + 3y)^5$ using Pascal's triangle.

Solution:

Step 1 — Fetch the row. Index 5: $1, 5, 10, 10, 5, 1$.

Step 2 — Attach falling and rising powers. $(2x)^5 + 5(2x)^4(3y) + 10(2x)^3(3y)^2 + 10(2x)^2(3y)^3 + 5(2x)(3y)^4 + (3y)^5$.

Step 3 — Evaluate each term. $(2x)^5 = 32x^5$; $5 \cdot 16 \cdot 3 = 240$; $10 \cdot 8 \cdot 9 = 720$; $10 \cdot 4 \cdot 27 = 1080$; $5 \cdot 2 \cdot 81 = 810$; $3^5 = 243$.

$$= 32x^5 + 240x^4y + 720x^3y^2 + 1080x^2y^3 + 810xy^4 + 243y^5$$

Step 4 — Check. At $x = y = 1$: $32+240+720+1080+810+243 = 3125 = 5^5 = (2+3)^5$ ✓.

Takeaway: The whole quantity $2x$ carries each power — and the $x=y=1$ substitution is a free full-expansion check.

Example 4: A quick cube

Expand $(x + 2y)^3$ by triangle row.

Solution:

Step 1 — Row 3. $1, 3, 3, 1$.

Step 2 — Expand. $x^3 + 3x^2(2y) + 3x(2y)^2 + (2y)^3 = x^3 + 6x^2y + 12xy^2 + 8y^3$.

Step 3 — Check. At $x = y = 1$: $1 + 6 + 12 + 8 = 27 = 3^3$ ✓.

Takeaway: $(2y)^2 = 4y^2$ and $(2y)^3 = 8y^3$ — the inner coefficient compounds.

Example 5: Spot the error

A student writes $(3a + b)^4 = 3a^4 + 4 \cdot 3a^3b + 6 \cdot 3a^2b^2 + 4 \cdot 3ab^3 + b^4$. Correct the mistake.

Solution:

Step 1 — Identify the error. The student treated $(3a)^k$ as $3a^k$; the base $3a$ must carry each power: $(3a)^4 = 81a^4$, $(3a)^3 = 27a^3$, $(3a)^2 = 9a^2$.

Step 2 — Rebuild. $81a^4 + 4(27a^3)b + 6(9a^2)b^2 + 4(3a)b^3 + b^4$.

Step 3 — Simplify. $81a^4 + 108a^3b + 54a^2b^2 + 12ab^3 + b^4$.

Step 4 — Check. At $a = b = 1$: $81 + 108 + 54 + 12 + 1 = 256 = 4^4 = (3+1)^4$ ✓.

Takeaway: Whenever the binomial's letters have coefficients, bracket them before raising.