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A question asks for C₁ + 2C₂ + 3C₃ + … + nCₙ, and adding the terms one by one is hopeless for n = 20. Every sum of this kind is hiding a single expansion, (1 + x)ⁿ, and a single move that turns it into the sum you want. This post gives the four moves, works an example of each with checked numbers and finishes with the mistakes and a practice routine. Here Cᵣ means ⁿCᵣ.
Where is this in the syllabus?
The NCERT Class 11 chapter Binomial Theorem (reprint 2026-27) derives two sums in section 7.2.2: 2ⁿ = C₀ + C₁ + … + Cₙ, and 0 = C₀ − C₁ + C₂ − … + (−1)ⁿCₙ. The JEE (Main) 2026 syllabus lists the binomial theorem for a positive integral index, "general term and middle term and simple applications". The JEE (Advanced) 2026 brochure goes further and names "properties of binomial coefficients". The sums below beyond the first two are not in NCERT; they are standard JEE Advanced material. Check the current year's documents.
Move 1: put x = 1 or x = −1
Start from (1 + x)ⁿ = C₀ + C₁x + C₂x² + … + Cₙxⁿ.
x = 1 gives C₀ + C₁ + … + Cₙ = 2ⁿ.
x = −1 gives C₀ − C₁ + C₂ − … = 0.
Add the two lines and divide by 2: the even-position coefficients C₀ + C₂ + C₄ + … sum to 2ⁿ⁻¹. Subtract and divide by 2: the odd ones C₁ + C₃ + … also sum to 2ⁿ⁻¹.
Check, n = 6. The row is 1, 6, 15, 20, 15, 6, 1. The total is 64 = 2⁶. The even positions give 1 + 15 + 15 + 1 = 32, and the odd positions 6 + 20 + 6 = 32. Each is 2⁵.
Move 2: differentiate
Differentiate (1 + x)ⁿ with respect to x:
n(1 + x)ⁿ⁻¹ = C₁ + 2C₂x + 3C₃x² + … + nCₙxⁿ⁻¹.
Now put x = 1: C₁ + 2C₂ + 3C₃ + … + nCₙ = n·2ⁿ⁻¹.
Example, n = 10. The sum is 10 × 2⁹ = 5120. A direct check by adding r × ¹⁰Cᵣ for r = 0 to 10 gives 5120.
There is also a no-calculus route: r × ⁿCᵣ = n × ⁿ⁻¹Cᵣ₋₁, so the sum becomes n times the sum of a full row of n − 1, which is n·2ⁿ⁻¹. Differentiating a second time gives Σ r(r − 1)Cᵣ = n(n − 1)·2ⁿ⁻², and adding Σ rCᵣ to it gives Σ r²Cᵣ = n(n + 1)·2ⁿ⁻². For n = 6 that is 6 × 7 × 16 = 672.
Move 3: integrate
Integrate both sides of (1 + x)ⁿ from 0 to 1:
[(1 + x)ⁿ⁺¹ ÷ (n + 1)] from 0 to 1 = C₀ + C₁/2 + C₂/3 + … + Cₙ/(n + 1).
So C₀ + C₁/2 + C₂/3 + … + Cₙ/(n + 1) = (2ⁿ⁺¹ − 1) ÷ (n + 1).
Example, n = 6. The sum is (128 − 1) ÷ 7 = 127/7. Adding 1 + 6/2 + 15/3 + 20/4 + 15/5 + 6/6 + 1/7 = 1 + 3 + 5 + 5 + 3 + 1 + 1/7 = 18 + 1/7 = 127/7. Correct.
Integrate from −1 to 0 instead and you get the alternating version, C₀ − C₁/2 + C₂/3 − … = 1/(n + 1). For n = 6 that is 1/7.
Move 4: multiply two expansions
Write (1 + x)ⁿ(1 + x)ⁿ = (1 + x)²ⁿ. On the left, the coefficient of xⁿ comes from pairing Cᵣ in the first bracket with Cₙ₋ᵣ in the second. Since Cₙ₋ᵣ = Cᵣ, that coefficient is C₀² + C₁² + … + Cₙ². On the right it is ²ⁿCₙ.
So C₀² + C₁² + … + Cₙ² = ²ⁿCₙ.
Example, n = 6. 1 + 36 + 225 + 400 + 225 + 36 + 1 = 924, and ¹²C₆ = 924. Correct.
The same trick with the coefficient of xⁿ⁻¹ gives C₀C₁ + C₁C₂ + … + Cₙ₋₁Cₙ = ²ⁿCₙ₋₁. For n = 6 that is ¹²C₅ = 792.
Which mistakes cost marks?
Starting the sum at the wrong term. C₁ + C₂ + … + Cₙ is 2ⁿ − 1, not 2ⁿ.
Forgetting to put x = 1 after differentiating, or putting it in before.
Dropping the lower limit when integrating. The constant from x = 0 is why the answer has 2ⁿ⁺¹ − 1, not 2ⁿ⁺¹.
Using Move 1 when the coefficients carry weights such as r or 1/(r + 1).
Not checking. A small case, n = 2 or n = 3, takes thirty seconds and catches most slips.
For coefficients of a particular term and middle terms, see our post on the binomial theorem for JEE, and for remainder questions, binomial remainders and divisibility.
If you can see the answer but not which move gets you there, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
A practice routine
Write rows n = 4, 5 and 6 of Pascal's triangle and verify every identity in this post on them.
For each of the four moves, do two problems with fresh values of n.
Mix them: shuffle eight sums and, before solving, write which move each needs.
Check every answer with n = 2.
Today, find C₁ + 2C₂ + … + 8C₈ for n = 8 two ways, by differentiating and by r × ⁿCᵣ = n × ⁿ⁻¹Cᵣ₋₁.
