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The sum 1·nC1 + 2·nC2 + … + n·nCn has a closed form: n·2ⁿ⁻¹. ProofWiki's entry on this identity gives three separate proofs, two of them below, and the Wikipedia article on binomial coefficients lists it under "Sums of the binomial coefficients". Learn both routes, because each one takes you to different problems.
Before any algebra, check a small case. For n = 4 the coefficients are 1, 4, 6, 4, 1. Multiply each by r and you get 0 + 4 + 12 + 12 + 4 = 32, and 4·2³ = 32.
How do you get Σ r·nCr by differentiating (1 + x)ⁿ?
Start from the binomial theorem:
(1 + x)ⁿ = nC0 + nC1·x + nC2·x² + … + nCn·xⁿ
Differentiate both sides with respect to x. On the left you get n(1 + x)ⁿ⁻¹. On the right each term nCr·xʳ becomes r·nCr·xʳ⁻¹, and the constant nC0 drops out:
n(1 + x)ⁿ⁻¹ = 1·nC1 + 2·nC2·x + 3·nC3·x² + … + n·nCn·xⁿ⁻¹
Now put x = 1. Every power of x becomes 1, so the right side is exactly the sum you want and the left side is n·2ⁿ⁻¹.
The strength of this route is that x is a dial you can turn. Put x = −1 instead and, for n ≥ 2, the left side is 0, so 1·nC1 − 2·nC2 + 3·nC3 − … = 0. One derivative gives you two results.
Can you prove Σ r·nCr = n·2ⁿ⁻¹ without calculus?
Yes. The key step is r·nCr = n·(n−1)C(r−1), an identity that the same Wikipedia article lists in the equivalent form nCr = (n/r)·(n−1)C(r−1), and that the first proof on ProofWiki uses. You can check it from factorials:
r·nCr = r·n!/(r!(n−r)!) = n!/((r−1)!(n−r)!) = n·(n−1)!/((r−1)!(n−r)!) = n·(n−1)C(r−1)
Now add from r = 1 to n. The r = 0 term is zero anyway, so you lose nothing by starting at 1:
Σ r·nCr = n·[(n−1)C0 + (n−1)C1 + … + (n−1)C(n−1)] = n·2ⁿ⁻¹
The bracket is the whole of row n−1 of Pascal's triangle, which adds up to 2ⁿ⁻¹.
There is a counting story too. From n students, pick a team of any size and then a captain from the team. Counting team first: choose r members in nCr ways, then the captain in r ways, giving Σ r·nCr. Counting captain first: n choices of captain, then each of the other n − 1 students is either in or out, giving 2ⁿ⁻¹. Both count the same teams, so the two answers are equal.
How do you find Σ r²·nCr?
The common trap: differentiating twice gives Σ r(r−1)·nCr·xʳ⁻², not Σ r²·nCr·xʳ⁻². Two fixes work.
Fix 1: split r². Write r² = r(r−1) + r. Differentiating (1 + x)ⁿ twice and putting x = 1 gives Σ r(r−1)·nCr = n(n−1)·2ⁿ⁻². The factorial route agrees, because r(r−1)·nCr = n(n−1)·(n−2)C(r−2). Add the first sum:
Σ r²·nCr = n(n−1)·2ⁿ⁻² + n·2ⁿ⁻¹ = n·2ⁿ⁻²·(n − 1 + 2) = n(n+1)·2ⁿ⁻²
Fix 2: multiply by x before the second derivative. From n(1 + x)ⁿ⁻¹ = Σ r·nCr·xʳ⁻¹, multiply both sides by x to get nx(1 + x)ⁿ⁻¹ = Σ r·nCr·xʳ. Differentiating now brings down another r:
n(1 + x)ⁿ⁻¹ + n(n−1)x(1 + x)ⁿ⁻² = Σ r²·nCr·xʳ⁻¹
At x = 1 this is again n·2ⁿ⁻¹ + n(n−1)·2ⁿ⁻² = n(n+1)·2ⁿ⁻².
This matches the formula (n + n²)·2ⁿ⁻² in the Wikipedia binomial coefficient article. Check with n = 4: 0 + 4 + 24 + 36 + 16 = 80, and 4·5·2² = 80.
Which route should you use in an exam?
The weight is r, r(r−1) or r(r−1)(r−2): use the factorial identity. It is quick and leaves little room for sign slips.
The weight is r² or r³: split it into those falling products (r² = r(r−1) + r), or use the multiply-by-x trick.
The signs alternate, or there is an extra factor like 2ʳ: differentiate, then choose x (x = −1, x = 2) to match.
Before you write the final answer, test n = 2 or n = 3 by hand.
The same identity makes the mean of a binomial distribution come out as np; we walk through that in why the binomial mean is np and the variance npq. For the plain sums these build on, see our guide to sums of binomial coefficients, and for the identity toolkit, nCr identities and Pascal's rule.
If you want to practise, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer, so you can see why a problem calls for differentiation or for the factorial identity. If Pascal's triangle row sums feel shaky, start with the mathematics lessons in the Fermi Concept Library.
