Binomial coefficients and partial sums

Binomial coefficients appear throughout the HSC Extension 1 course, particularly in the binomial theorem and combinatorial identities. In this post we work through a single, carefully chosen sequence of results that shows how powerful the binomial expansion can be: by substituting special values and differentiating, you can generate sum identities almost effortlessly. Mastering these techniques will strengthen your ability to manipulate series and to recognise hidden relationships in exam problems.

Problem Statement

Let

$$S_n=\sum_{k=0}^n \binom{n}{k}$$

be the sum of the binomial coefficients for a fixed integer $n\ge1$.

• Use the binomial theorem expansion of $(1+x)^n$ to prove that $S_n=2^n$.
• Use Pascal's identity

$$\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$$

to prove the recurrence relation $S_n=2S_{n-1}$.

• Evaluate the alternating partial sum

$$A_n=\sum_{k=0}^n (-1)^k\binom{n}{k}$$

and explain why the sum of even-indexed coefficients equals the sum of odd-indexed coefficients.

• By differentiating the expansion of $(1+x)^n$ with respect to $x$, show that

$$\sum_{k=1}^n k\binom{n}{k}=n2^{n-1}.$$

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Hints

• For (i): Substitute $x=1$ into the binomial theorem.
• For (ii): Sum Pascal's identity over the appropriate range of $k$.
• For (iii): Substitute $x=-1$ into the binomial theorem.
• For (iv): Differentiate first, then substitute $x=1$.

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Solutions

(i) We start with the binomial theorem expansion of $(1+x)^n$, which expresses the power as a sum of binomial coefficients weighted by powers of $x$. Substituting $x=1$ makes the right‑hand side simply the sum of all coefficients $S_n$, while the left‑hand side becomes $2^n$.

By the binomial theorem,

$$(1+x)^n=\sum_{k=0}^n \binom{n}{k}x^k.$$

Putting $x=1$ gives

$$2^n=\sum_{k=0}^n \binom{n}{k}=S_n.$$

(ii) Pascal's identity allows us to write each $\binom{n}{k}$ in terms of two coefficients from the previous row. Summing this identity over all $k$ splits the total sum into two parts, each of which is essentially $S_{n-1}$ once we account for boundary terms.

Using Pascal's identity,

$$\begin{aligned} S_n &=\sum_{k=0}^n\binom{n}{k}\\ &=\sum_{k=0}^n\binom{n-1}{k}+\sum_{k=0}^n\binom{n-1}{k-1}. \end{aligned}$$

The out‑of‑range terms are zero, so each sum equals $S_{n-1}$. Hence

$$S_n=2S_{n-1}.$$

(iii) To evaluate the alternating sum we substitute $x=-1$ into the binomial expansion. This cancels terms pairwise and gives a neat zero, revealing a balance between even‑indexed and odd‑indexed coefficients.

By the binomial theorem again,

$$A_n=\sum_{k=0}^n(-1)^k\binom{n}{k}=(1-1)^n=0.$$

Therefore

$$\sum_{\substack{0\le k\le n\\ k\text{ even}}}\binom{n}{k} - \sum_{\substack{0\le k\le n\\ k\text{ odd}}}\binom{n}{k} =0,$$

so the even‑indexed and odd‑indexed sums are equal.

(iv) Differentiating the expansion with respect to $x$ brings down a factor of $k$ from each term, immediately turning a sum of coefficients into a weighted sum. Then we simply set $x=1$ to obtain a closed form.

Differentiate

$$(1+x)^n=\sum_{k=0}^n\binom{n}{k}x^k$$

to get

$$n(1+x)^{n-1}=\sum_{k=1}^n k\binom{n}{k}x^{k-1}.$$

Putting $x=1$ gives

$$\sum_{k=1}^n k\binom{n}{k}=n2^{n-1}.$$

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Takeaways

• Substituting special values into the binomial theorem turns an expansion into a sum identity.
• Pascal's identity explains why the total sum of coefficients doubles from one row to the next.
• Differentiating a generating expansion is a powerful way to evaluate weighted sums.

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Further Readings

HSC Sequences, HSC Induction, HSC Proofs, HSC Complex Numbers