In short

You already met the trap at n = 2: (a+b)^2 \neq a^2 + b^2. That was bad enough — you missed one cross term, 2ab. Now scale the same wrong instinct up to (a+b)^n and it becomes catastrophic. At n = 3 you miss two whole terms. At n = 4 you miss three. At n = 10 you miss nine terms, and the gap between the wrong answer and the right one grows so fast that the wrong answer becomes a microscopic fraction of the truth. The actual rule is the Binomial Theorem: (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k. Every row of Pascal's triangle is the antidote. This is the most-corrected-and-still-repeated identity error in JEE Mains, and the reason is the same as the squared case — your pattern-matcher confuses an exponent with a multiplier — but the consequences are an order of magnitude worse. Why: the missed terms aren't a fixed correction; they multiply combinatorially with n. By n = 4 the wrong answer is already off by more than 80\% on simple integer inputs.

You met this disease in the squared form. The cure for (a+b)^2 is in (a+b)² = a² + b², and the subtraction-side cousin lives in (a−b)² = a² − b². This article is the general version. Same trap, scaled to any exponent n. And the moment n grows past 2, the trap stops being a two-mark slip and starts being a problem-destroying disaster.

Walk into a JEE Mains coaching staffroom in Kota and ask which algebra mistake students keep repeating despite being told a hundred times. The answer is this one: (a+b)^n written as a^n + b^n. Not in class 9, not in class 10 — at JEE level, on Binomial Theorem questions, on Limits questions, on Sequences questions. The little exponent is a multiplier in the brain's pattern-cache, and the cache is wrong.

The trap, generalised

At n = 2, the wrong identity skips 2ab. One missing term.

At n = 3, the wrong identity skips 3a^2 b + 3ab^2. Two missing terms.

At n = 4, the wrong identity skips 4a^3 b + 6a^2 b^2 + 4ab^3. Three missing terms.

At n = 10, the wrong identity skips nine terms — every term except the first and the last. Why: (a+b)^n has exactly n+1 terms in its expansion (one for each row of Pascal's triangle), but the wrong shortcut a^n + b^n keeps only the two extreme terms. Every middle term is lost.

The truth is the binomial theorem:

(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} \, a^{n-k} \, b^k

where \binom{n}{k} = \dfrac{n!}{k!(n-k)!} is the binomial coefficient — the count of ways to choose k of the n brackets to contribute a b instead of an a. The coefficients are exactly the rows of Pascal's triangle:

n = 0:                1
n = 1:              1   1
n = 2:            1   2   1
n = 3:          1   3   3   1
n = 4:        1   4   6   4   1
n = 5:      1   5   10  10   5   1

Only the first and last entries of each row are 1. Every other coefficient is \geq 2 — and those are exactly the cross-terms the wrong shortcut throws away.

The trap is not just bigger — it is exponentially bigger

Bar chart showing how the gap between (2+3) to the n and 2 to the n plus 3 to the n grows with nA bar chart for n = 2, 3, 4, 5. For each n, two bars: a tall blue bar showing the correct value of (2+3) to the n, and a much shorter orange bar showing the wrong shortcut 2 to the n plus 3 to the n. At n=2 the bars are 25 vs 13. At n=3 the bars are 125 vs 35. At n=4 the bars are 625 vs 97. At n=5 the bars are 3125 vs 275. The orange bars stay tiny while the blue bars race upward. Trap escalation: (2+3)ⁿ vs 2ⁿ + 3ⁿ 0 800 1600 2400 3200 n = 2 25 13 n = 3 125 35 n = 4 625 97 n = 5 3125 275 truth (2+3)ⁿ wrong 2ⁿ+3ⁿ
For $a = 2, b = 3$, the truth $(2+3)^n = 5^n$ races upward while the wrong shortcut $2^n + 3^n$ creeps along the floor. At $n = 5$ the truth is $3125$ while the shortcut gives $275$ — the wrong answer is less than $9\%$ of the right one. By $n = 10$ the truth is $9{,}765{,}625$ and the shortcut is $1{,}083$ — the wrong answer is $0.011\%$ of reality. This is what "combinatorial" looks like on a bar chart.

The chart is the article in one picture. The blue bars are reality. The orange bars are what your hand wants to write. The gap between them is not a fixed mistake — it is a chasm that opens wider with every increase in n.

Three worked examples

Example 1 — $n = 3$: the cube of a sum is *not* the sum of cubes

The truth, from row 3 of Pascal's triangle (1, 3, 3, 1):

(a + b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3

The wrong shortcut:

(a + b)^3 \stackrel{?}{=} a^3 + b^3

The two missing terms are 3a^2 b + 3ab^2. Together they equal 3ab(a+b), which is exactly what makes (a+b)^3 different from a^3 + b^3.

Why: a^3 + b^3 is a real and beautiful expression in its own right — it factors as (a+b)(a^2 - ab + b^2) — but it is not the same object as (a+b)^3. They are two distinct algebraic creatures. Confusing them is like confusing a square with a square root: same letters in the symbol, completely different meanings.

Test with a = 1, b = 2.

  • Truth: (1+2)^3 = 3^3 = 27.
  • Wrong shortcut: 1^3 + 2^3 = 1 + 8 = 9.

You are off by 18. The missing pieces 3a^2 b + 3ab^2 = 3(1)(2)(1+2) = 18. The arithmetic exposes the bug.

And there is a separate, often-confused identity: a^3 + b^3 = (a+b)(a^2 - ab + b^2). That identity is true, and it is the sum of cubes factorisation. It is the inverse direction — given a^3 + b^3, find its factors. But (a+b)^3 and a^3 + b^3 are different expressions: the first equals 27, the second equals 9, when a = 1, b = 2. They cannot be equal in general.

Example 2 — $n = 4$: five terms, only two are pure powers

Pascal's triangle, row 4: 1, 4, 6, 4, 1. So:

(a + b)^4 = a^4 + 4a^3 b + 6a^2 b^2 + 4ab^3 + b^4

Five terms. Three of them are pure cross-terms — terms with both a and b in them. Only the very first (a^4) and the very last (b^4) are pure powers of one variable. Everything in the middle is a mixture, and that middle is most of the answer.

Why: think of (a+b)^4 as (a+b)(a+b)(a+b)(a+b) — four brackets multiplied together. To form a term in the expansion, you pick one item (either a or b) from each bracket. There are 2^4 = 16 ways to do this, and they sort by how many b's you picked: zero b's (only one way, gives a^4), one b (four ways — Pascal's 4 — gives 4a^3 b), two b's (six ways — Pascal's 6 — gives 6a^2 b^2), and so on. The cross-terms are not "extra"; they are most of the answer.

Test with a = 1, b = 1 — the laziest possible test.

  • Truth: (1+1)^4 = 2^4 = 16.
  • Wrong shortcut: 1^4 + 1^4 = 2.

The wrong answer is one-eighth of the right one. Eight times too small. The missing 4 + 6 + 4 = 14 comes from the three middle terms.

If you ever see your hand write (a+b)^4 = a^4 + b^4 in a JEE Mains problem, this is the test that should fire automatically: plug in a = b = 1 and notice that 16 \neq 2.

Example 3 — Numeric: $(2+3)^4$ is not $2^4 + 3^4$

Sometimes the cleanest cure is pure arithmetic. No letters. Just numbers.

Truth:

(2 + 3)^4 = 5^4 = 625

Wrong shortcut:

2^4 + 3^4 = 16 + 81 = 97

The gap is 625 - 97 = 528. The wrong answer captures only 97/625 \approx 15.5\% of the truth — less than one-sixth. Why: the missing 528 comes from the three middle terms: 4 \cdot 2^3 \cdot 3 + 6 \cdot 2^2 \cdot 3^2 + 4 \cdot 2 \cdot 3^3 = 96 + 216 + 216 = 528. The middle terms aren't a small correction — they are the calculation.

This is the test you can run on any squared/cubed/quartic bracket in three seconds: pick a = 2 and b = 3 (or any two small unequal numbers), compute both sides, and look at the ratio. If the ratio is far from 1, your shortcut is wrong by a country mile.

Why this trap survives even after teachers warn

You learnt the distributive law in class 6 or 7:

k(a + b) = ka + kb

This is true. It is one of the most reliable rules in arithmetic. You have used it thousands of times and it has never let you down. Multiply by 3? Push the 3 inside. Multiply by \tfrac{1}{2}? Push the \tfrac{1}{2} inside. The pattern is rock-solid.

Then you meet (a+b)^k. The little k outside the bracket looks like a multiplier. Your pattern-matcher fires. Push it inside:

(a+b)^k \stackrel{?}{=} a^k + b^k

It feels just like the distributive law. Same shape on the page. Same little superscript or coefficient outside the bracket. Same urge to "drop it onto each term".

But the two operations are fundamentally different. Why: k \cdot (a+b) adds the bracket to itself k times — which is repeated addition, and addition distributes. (a+b)^k multiplies the bracket by itself k times — which is repeated multiplication, and multiplication does not distribute over addition the same way. When you multiply two sums, every term in the first sum has to meet every term in the second; cross-terms appear. Repeated multiplication makes the cross-term explosion bigger and bigger.

The fix is to separate the two patterns in your head before the exam, not during. Teach yourself:

The visual difference is one symbol — 3 \cdot (a+b) vs (a+b)^3 — but the algebra on the right side is a different universe. Train your eye to feel the difference.

This is also why the trap survives JEE coaching: students who rote-memorise the binomial theorem still default to the wrong shortcut under exam pressure, because the habit of "push the outside thing in" was trained over six years of distributive-law practice. Killing the habit takes deliberate counter-training: every time you see an exponent outside a bracket, pause, write "Pascal's row n", then expand. After a few weeks of this, the wrong shortcut stops appearing.

The number-test reflex

For any n, the fastest bug-detector is to test with small numbers. Pick a = 1, b = 1 — the laziest possible test.

So the wrong shortcut gives you 2 no matter what n is, while the truth doubles with every increase in n. At n = 10, the truth is 1024 and the shortcut is 2. The gap is so absurd that you can never write the wrong identity again with a straight face.

Make this the reflex: any time your hand writes (a+b)^n = \text{something}, plug in a = b = 1 and check that "something" gives you 2^n. If it gives you 2, your line is wrong. Three seconds. Eight marks saved on a JEE problem.

Putting it together

The squared-form misconception killer gave you three habits — number test, rectangle picture, FOIL out loud. For general n, the rectangle picture becomes a 2 \times 2 \times \cdots \times 2 hyper-grid (one factor for each bracket), and FOIL becomes "enumerate 2^n products". Both still work, but the easier mental discipline at higher n is:

Do this for two weeks of practice problems and the wrong shortcut a^n + b^n stops appearing in your work — including under the time pressure of the JEE Mains paper, where this mistake costs more marks across more chapters than almost any other algebra slip.

References

  1. Wikipedia: Binomial theorem — the full statement, history, and proof of (a+b)^n = \sum \binom{n}{k} a^{n-k} b^k.
  2. Wikipedia: Freshman's dream — the formal name for the false identity (a+b)^n = a^n + b^n over the real numbers.
  3. Wikipedia: Pascal's triangle — the array of binomial coefficients, with the Indian history through Pingala and Halayudha.
  4. NCERT Class 11 Mathematics, Chapter 8: Binomial Theorem — the standard JEE-aligned treatment, with worked problems on (a+b)^n.
  5. Khan Academy: Binomial theorem — short videos with examples for n = 2 through n = 6.