In short

When you add the expansions of (x + y)^n and (x - y)^n, every odd-power term cancels: you get twice the sum of the even-indexed terms. When you subtract them, every even-power term cancels: you get twice the sum of the odd-indexed terms. These two identities — (x+y)^n + (x-y)^n and (x+y)^n - (x-y)^n — let you isolate half the terms of a binomial expansion, which is exactly what many JEE and competition problems demand.

Take a specific example: (1 + \sqrt{2})^6 + (1 - \sqrt{2})^6. If you expand both separately, each has seven terms with messy powers of \sqrt{2}. But add them, and every term with (\sqrt{2})^1, (\sqrt{2})^3, or (\sqrt{2})^5 — the odd powers — cancels perfectly. Only the rational terms survive.

That is the power of special expansions: they turn a pair of complicated expressions into something clean.

The core identities

Start with the binomial theorem. For any positive integer n:

(x + y)^n = \sum_{r=0}^{n} \binom{n}{r}\, x^{n-r}\, y^r
(x - y)^n = \sum_{r=0}^{n} \binom{n}{r}\, x^{n-r}\, (-y)^r = \sum_{r=0}^{n} (-1)^r \binom{n}{r}\, x^{n-r}\, y^r

The only difference between the two expansions is the factor (-1)^r. When r is even, (-1)^r = 1 and the terms match. When r is odd, (-1)^r = -1 and the terms have opposite signs.

Terms of (x+y) to the n and (x-y) to the n comparedTwo rows of terms. Top row shows the expansion of (x+y) to the n: all terms with plus signs. Bottom row shows (x-y) to the n: even-indexed terms positive, odd-indexed terms negative. Adding the two rows doubles the even terms and cancels the odd terms. Comparing (x+y)ⁿ and (x−y)ⁿ term by term (x+y)ⁿ: +T₁ +T₂ +T₃ +T₄ +T₅ ... (x−y)ⁿ: +T₁ −T₂ +T₃ −T₄ +T₅ ... Add: 2T₁ 0 2T₃ 0 2T₅ ... Subtract: 0 2T₂ 0 2T₄ 0 ... T₁, T₃, T₅, ... have even r (r = 0, 2, 4, ...) — these are the "even-indexed" terms T₂, T₄, T₆, ... have odd r (r = 1, 3, 5, ...) — these are the "odd-indexed" terms
When you add the two expansions, odd-powered terms cancel (because $+T + (-T) = 0$). When you subtract, even-powered terms cancel (because $+T - (+T) = 0$). The surviving terms are doubled.

The formal statements

Special expansion identities

For any positive integer n and any x, y:

Sum:

(x+y)^n + (x-y)^n = 2\sum_{\substack{r=0 \\ r\text{ even}}}^{n} \binom{n}{r}\, x^{n-r}\, y^r

Difference:

(x+y)^n - (x-y)^n = 2\sum_{\substack{r=1 \\ r\text{ odd}}}^{n} \binom{n}{r}\, x^{n-r}\, y^r

For the common case x = 1, y = x:

(1+x)^n + (1-x)^n = 2\left[1 + \binom{n}{2}x^2 + \binom{n}{4}x^4 + \cdots\right]
(1+x)^n - (1-x)^n = 2\left[\binom{n}{1}x + \binom{n}{3}x^3 + \binom{n}{5}x^5 + \cdots\right]

Counting the terms

How many terms does each special expansion have? If n is even, say n = 2m, the even values of r are 0, 2, 4, \ldots, 2m — that is m + 1 terms in the sum and m terms in the difference. If n is odd, say n = 2m + 1, the even values are 0, 2, \ldots, 2m (m + 1 terms) and the odd values are 1, 3, \ldots, 2m+1 (m + 1 terms).

Number of terms in special expansionsA table showing: when n is even (n=2m), the sum has m+1 terms and the difference has m terms. When n is odd (n=2m+1), both the sum and the difference have m+1 terms. Number of surviving terms Sum (even r) Difference (odd r) n even (n=2m) m + 1 terms m terms n odd (n=2m+1) m + 1 terms m + 1 terms Example: n = 6 (even, m = 3) → sum has 4 terms, difference has 3 terms Example: n = 7 (odd, m = 3) → sum has 4 terms, difference has 4 terms
The total number of terms in the original expansion is $n + 1$. The special expansions split these roughly in half. When $n$ is even, the sum gets one more term than the difference.

A concrete expansion: n = 6

Take (x + y)^6 and (x - y)^6. The full expansions are:

(x+y)^6 = x^6 + 6x^5y + 15x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6xy^5 + y^6
(x-y)^6 = x^6 - 6x^5y + 15x^4y^2 - 20x^3y^3 + 15x^2y^4 - 6xy^5 + y^6

Add them:

(x+y)^6 + (x-y)^6 = 2(x^6 + 15x^4y^2 + 15x^2y^4 + y^6)

Four terms survive — those with y^0, y^2, y^4, y^6 (even powers of y).

Subtract:

(x+y)^6 - (x-y)^6 = 2(6x^5y + 20x^3y^3 + 6xy^5)

Three terms survive — those with y^1, y^3, y^5 (odd powers of y).

Special expansions for n equals 6 with terms highlightedThe seven terms of (x+y) to the 6 are listed. Even-indexed terms (r=0,2,4,6) are highlighted in red and appear in the sum identity. Odd-indexed terms (r=1,3,5) appear in the difference identity. (x+y)⁶ — which terms survive in each identity? x⁶ r=0 6x⁵y r=1 15x⁴y² r=2 20x³y³ r=3 15x²y⁴ r=4 6xy⁵ r=5 y⁶ r=6 Sum: 2(x⁶ + 15x⁴y² + 15x²y⁴ + y⁶) even r terms survive (red terms above) Difference: 2(6x⁵y + 20x³y³ + 6xy⁵) odd r terms survive (uncoloured terms above) Check: sum of all 7 coefficients: 1+6+15+20+15+6+1 = 64 = 2⁶ ✓
The seven terms of $(x+y)^6$ split into two groups. The even-$r$ terms (shown in red) form the sum identity; the odd-$r$ terms form the difference identity. Each group is multiplied by $2$ in the result.

The (1+x)^n versions and their coefficient sums

Setting x = 1 and y = x in the identities gives the forms you will use most often.

Sum of even-indexed binomial coefficients. Set x = 1 in (1+x)^n + (1-x)^n = 2[\text{even terms}]:

(1+1)^n + (1-1)^n = 2\left[\binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \cdots\right]
2^n + 0 = 2\left[\binom{n}{0} + \binom{n}{2} + \cdots\right]
\binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \cdots = 2^{n-1}

Sum of odd-indexed binomial coefficients. Similarly:

\binom{n}{1} + \binom{n}{3} + \binom{n}{5} + \cdots = 2^{n-1}

The even-indexed and odd-indexed coefficients each sum to exactly half of 2^n. This is a beautiful consequence: out of the 2^n subsets of an n-element set, exactly half have even size and half have odd size.

Even and odd binomial coefficients each sum to 2 to the n-1Pascal's triangle row for n=5 is shown: 1, 5, 10, 10, 5, 1. The even-indexed coefficients 1, 10, 5 are highlighted and sum to 16. The odd-indexed coefficients 5, 10, 1 also sum to 16. Both equal 2 to the 4 = 16. n = 5: even vs odd coefficient sums 1 C(5,0) 5 C(5,1) 10 C(5,2) 10 C(5,3) 5 C(5,4) 1 C(5,5) Even r: 1 + 10 + 5 = 16 Odd r: 5 + 10 + 1 = 16 Both = 2⁴ = 16 = half of 2⁵ = 32 Half the subsets of {1,2,3,4,5} have even size, half have odd size
For $n = 5$, the row of Pascal's triangle is $1, 5, 10, 10, 5, 1$, summing to $32$. The even-indexed coefficients sum to $16$, and so do the odd-indexed ones — each half equals $2^{n-1} = 16$.

Interactive: toggle between sum and difference

Drag the red point to choose n from 2 to 8. The figure shows the coefficients of (1+x)^n, with even-indexed terms in red and odd-indexed terms in gray. The running totals at the bottom confirm that both groups always sum to 2^{n-1}.

Interactive: even and odd binomial coefficient sumsA bar chart of binomial coefficients C(n,r) for r from 0 to n. Even-r bars are coloured red, odd-r bars are gray. A draggable point controls n. Two running totals at the bottom show that even and odd sums are always equal. r C(n,r) ↔ drag the red point
For every $n$, the even-indexed coefficients (red) and the odd-indexed coefficients (gray) sum to the same value: $2^{n-1}$. Drag to see this hold for different $n$.

Applications of the special expansion identities

Evaluating irrational expressions

The most common application: compute (a + \sqrt{b})^n + (a - \sqrt{b})^n.

Since (a + \sqrt{b})^n + (a - \sqrt{b})^n contains only even powers of \sqrt{b}, every term is rational (because (\sqrt{b})^{2k} = b^k). All the irrational parts cancel.

Example. Compute (1 + \sqrt{3})^4 + (1 - \sqrt{3})^4.

Apply the sum formula with x = 1, y = \sqrt{3}, n = 4:

= 2\left[\binom{4}{0} \cdot 1^4 + \binom{4}{2} \cdot 1^2 \cdot 3 + \binom{4}{4} \cdot 3^2\right]
= 2\left[1 + 6 \times 3 + 1 \times 9\right] = 2[1 + 18 + 9] = 2 \times 28 = 56

No surds anywhere in the answer.

Proving that an expression is an integer

To show that (3 + \sqrt{5})^7 + (3 - \sqrt{5})^7 is an integer, just note that (3 - \sqrt{5})^7 is positive but very small (since 3 - \sqrt{5} \approx 0.76, and 0.76^7 \approx 0.14). The sum identity tells you the result contains only even powers of \sqrt{5}, which are integers. So the expression is a positive integer. You do not even need to compute it — just the identity is enough.

Finding the integer nearest to (a + \sqrt{b})^n

If 0 < a - \sqrt{b} < 1, then (a - \sqrt{b})^n is a small positive number less than 1. The sum (a + \sqrt{b})^n + (a - \sqrt{b})^n is an integer (by the special expansion identity). So (a + \sqrt{b})^n equals that integer minus a small positive fraction — the nearest integer is the sum itself.

Two worked examples

Example 1: Find the value of $(2 + \sqrt{3})^6 + (2 - \sqrt{3})^6$

Step 1. Identify x = 2, y = \sqrt{3}, n = 6. Apply the sum identity: only even powers of \sqrt{3} survive.

Why: (2 + \sqrt{3})^6 + (2 - \sqrt{3})^6 is the sum of conjugate expressions. By the special expansion, the odd-power terms cancel.

Step 2. List the surviving terms. Even values of r: 0, 2, 4, 6.

= 2\left[\binom{6}{0}\cdot 2^6 + \binom{6}{2}\cdot 2^4 \cdot (\sqrt{3})^2 + \binom{6}{4}\cdot 2^2 \cdot (\sqrt{3})^4 + \binom{6}{6}\cdot (\sqrt{3})^6\right]

Why: the sum identity gives twice the even-indexed terms. Each (\sqrt{3})^{2k} = 3^k, which is rational.

Step 3. Compute each term.

r = 0: 1 \cdot 64 = 64

r = 2: 15 \cdot 16 \cdot 3 = 720

r = 4: 15 \cdot 4 \cdot 9 = 540

r = 6: 1 \cdot 27 = 27

Why: \binom{6}{0} = 1, \binom{6}{2} = 15, \binom{6}{4} = 15, \binom{6}{6} = 1. And (\sqrt{3})^2 = 3, (\sqrt{3})^4 = 9, (\sqrt{3})^6 = 27.

Step 4. Add and double.

64 + 720 + 540 + 27 = 1351

Result: 2 \times 1351 = 2702.

Result. (2 + \sqrt{3})^6 + (2 - \sqrt{3})^6 = \mathbf{2702}.

Computing (2 + root 3) to the 6 plus (2 - root 3) to the 6A diagram showing the four surviving terms (r=0,2,4,6) and their values: 64, 720, 540, 27. The sum is 1351, doubled to give 2702. The odd-r terms are crossed out. (2+√3)⁶ + (2−√3)⁶ — even-r terms only r = 0 64 r = 2 720 r = 4 540 r = 6 27 Sum of four terms: 64 + 720 + 540 + 27 = 1351 Answer: 2 × 1351 = 2702 Check: (2+√3) ≈ 3.732, so (3.732)⁶ ≈ 2700.96. And (2−√3)⁶ ≈ (0.268)⁶ ≈ 0.0004. Sum ≈ 2700.96 + 0.0004 ≈ 2701. Close to 2702 — rounding confirms ✓
Only four terms (with even powers of $\sqrt{3}$) survive. Each $(\sqrt{3})^{2k}$ becomes $3^k$, so the result is a clean integer: $2702$. The numerical check using decimal approximations confirms the answer.

The key observation: (2 - \sqrt{3})^6 \approx 0.0004, which is negligible. So (2 + \sqrt{3})^6 \approx 2702 - 0.0004 \approx 2702. The nearest integer to (2 + \sqrt{3})^6 is 2702.

Example 2: Show that $(1 + \sqrt{2})^6 - (1 - \sqrt{2})^6 = 140\sqrt{2}$

Step 1. This is a difference of conjugates, so use the difference identity with x = 1, y = \sqrt{2}, n = 6. Only odd powers of \sqrt{2} survive.

Why: subtracting the two expansions cancels all even-power terms and doubles all odd-power terms.

Step 2. List the surviving terms. Odd values of r: 1, 3, 5.

= 2\left[\binom{6}{1}\cdot 1^5 \cdot (\sqrt{2})^1 + \binom{6}{3}\cdot 1^3 \cdot (\sqrt{2})^3 + \binom{6}{5}\cdot 1^1 \cdot (\sqrt{2})^5\right]

Why: the difference identity gives twice the odd-indexed terms. Each surviving term has an odd power of \sqrt{2}.

Step 3. Simplify each term.

r = 1: 6 \cdot \sqrt{2} = 6\sqrt{2}

r = 3: 20 \cdot 2\sqrt{2} = 40\sqrt{2}

r = 5: 6 \cdot 4\sqrt{2} = 24\sqrt{2}

Why: (\sqrt{2})^3 = 2\sqrt{2} and (\sqrt{2})^5 = 4\sqrt{2}. Every odd power of \sqrt{2} has \sqrt{2} as a factor.

Step 4. Combine.

= 2(6\sqrt{2} + 40\sqrt{2} + 24\sqrt{2}) = 2 \times 70\sqrt{2} = 140\sqrt{2}

Result. (1 + \sqrt{2})^6 - (1 - \sqrt{2})^6 = \mathbf{140\sqrt{2}}.

Computing (1+root 2) to the 6 minus (1-root 2) to the 6Three surviving terms with odd r values: r=1 gives 6 root 2, r=3 gives 40 root 2, r=5 gives 24 root 2. Sum is 70 root 2, doubled to 140 root 2. (1+√2)⁶ − (1−√2)⁶ — odd-r terms only r = 1: C(6,1)·(√2)¹ 6√2 r = 3: C(6,3)·(√2)³ 40√2 r = 5: C(6,5)·(√2)⁵ 24√2 Sum: (6 + 40 + 24)√2 = 70√2 Answer: 2 × 70√2 = 140√2 Check: (1+√2)⁶ ≈ (2.414)⁶ ≈ 197.99, (1−√2)⁶ ≈ (−0.414)⁶ ≈ 0.005 Difference ≈ 197.99. And 140√2 ≈ 197.99 ✓
Only three terms (with odd powers of $\sqrt{2}$) survive. Each has a factor of $\sqrt{2}$, so the entire expression is a rational multiple of $\sqrt{2}$. The decimal check confirms the answer.

Notice the pattern: the sum identity gives an integer, and the difference identity gives an integer times \sqrt{2}. This always happens when y = \sqrt{b} — the sum kills all irrationals, and the difference factors out exactly one \sqrt{b}.

Common confusions

Going deeper

If you can apply the sum and difference identities to compute conjugate-pair expressions and extract the coefficient sums 2^{n-1}, you have the core skill. The material below explores connections that the identities open up.

Connection to Chebyshev polynomials

The expression \frac{(x + \sqrt{x^2 - 1})^n + (x - \sqrt{x^2 - 1})^n}{2} is the Chebyshev polynomial of the first kind, T_n(x). It satisfies T_n(\cos\theta) = \cos(n\theta). The sum identity is what makes this polynomial rational in x — despite the \sqrt{x^2 - 1} in the definition, all irrational parts cancel.

Pell's equation and (a + b\sqrt{d})^n

The identity (a + b\sqrt{d})^n + (a - b\sqrt{d})^n = 2 \times \text{integer} is central to the theory of Pell's equation x^2 - dy^2 = 1. If (a, b) is a solution to a^2 - db^2 = 1, then the integer part of (a + b\sqrt{d})^n gives the n-th solution — and the special expansion identity is what guarantees that this integer part is well-defined.

Generating binomial coefficient sums

By substituting different values of x and y, you can generate many coefficient identities. Setting x = 1 and y = i (where i^2 = -1) gives:

(1 + i)^n + (1 - i)^n = 2\left[\binom{n}{0} - \binom{n}{2} + \binom{n}{4} - \cdots\right]

Since (1+i)^n = (\sqrt{2})^n \cdot e^{in\pi/4}, the left side equals 2^{n/2} \cdot 2\cos(n\pi/4), giving a closed form for the alternating even-indexed sum of binomial coefficients. These connections are explored further in binomial theorem with complex numbers.

Where this leads next