Applications of the Binomial Theorem
The binomial theorem is more than an expansion machine — it encodes deep combinatorial relationships. By substituting clever values of $x$ into $(1+x)^n$, you can evaluate entire sums of binomial coefficients in a single step, prove identities, and unlock connections between combinatorics and powers of 2. This lesson gives you the substitution toolkit.
The binomial theorem states $(1+x)^n = \,^nC_0 + \,^nC_1 x + \,^nC_2 x^2 + \cdots + \,^nC_n x^n$. Without substituting anything yet — what value of $x$ do you think would give you the sum $\,^nC_0 + \,^nC_1 + \,^nC_2 + \cdots + \,^nC_n$? What value gives the alternating sum $\,^nC_0 - \,^nC_1 + \,^nC_2 - \cdots$? Predict before reading on.
The entire lesson rests on two substitutions into $(1+x)^n = \sum_{r=0}^{n} \,^nC_r x^r$. Master them and almost all binomial-coefficient sum questions become one-liners.
Set $x = 1$ and every $x^r$ becomes 1, so the right side collapses to the plain sum of coefficients $= 2^n$. Set $x = -1$ and every $x^r$ alternates $\pm 1$, giving the alternating sum $= 0$. Any other value of $x$ is a "bonus substitution" that evaluates more exotic sums.
Key facts
- Substituting $x=1$: $\sum_{r=0}^n \,^nC_r = 2^n$
- Substituting $x=-1$: $\sum_{r=0}^n (-1)^r\,^nC_r = 0$
- Even and odd coefficient sums are each equal to $2^{n-1}$
Concepts
- Why substituting into both $x=1$ and $x=-1$ separates even and odd terms
- How the general term $\,^nC_r x^r$ becomes a specific coefficient sum for given $x$
- The connection between powers of 2 and subsets of an $n$-element set
Skills
- Evaluate sums of binomial coefficients using substitution
- Find sums weighted by powers of 2 or other constants
- Prove simple binomial coefficient identities using substitution
The binomial theorem gives us $(1+x)^n = \,^nC_0 + \,^nC_1 x + \,^nC_2 x^2 + \cdots + \,^nC_n x^n$.
Setting $x = 1$:
This says the sum of all entries in row $n$ of Pascal's triangle equals $2^n$. This also has a combinatorial meaning: $2^n$ is the number of subsets of an $n$-element set (each element is either included or not).
\,^nC_0 + \,^nC_1 + + \,^nC_n = 2^n (set x=1 in the binomial theorem); Combinatorial meaning: 2^n is the number of subsets of an n-element set
Pause — copy the $x=-1$ result into your book: $\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \cdots = 0$ (for $n \ge 1$); adding $x=1$ and $x=-1$ results: $2(\binom{n}{0} + \binom{n}{2} + \cdots) = 2^n$, so even-index sum $= 2^{n-1}$.
Quick check: Evaluate $\,^6C_0 + \,^6C_1 + \,^6C_2 + \,^6C_3 + \,^6C_4 + \,^6C_5 + \,^6C_6$.
We just saw that setting $x = 1$ in $(1+x)^n$ gives $\sum_{k=0}^n \binom{n}{k} = 2^n$ — the total count of subsets. That raises a question: setting $x = -1$ produces an alternating sum of binomial coefficients — what does it equal, and how does combining the $x=1$ and $x=-1$ results give the even-index and odd-index sums separately? This card answers it → $x = -1$ gives 0 (for $n \ge 1$); adding the two results isolates the even sum $= 2^{n-1}$.
Setting $x = -1$ in the binomial theorem:
This means the even-indexed coefficients and odd-indexed coefficients are equal. Adding both equations ($x=1$ and $x=-1$):
The even-position and odd-position coefficients each sum to half of $2^n$.
\,^nC_0 - \,^nC_1 + \,^nC_2 - = 0 (set x=-1 in the binomial theorem, n 1); Adding x=1 and x=-1 results: 2(\,^nC_0 + \,^nC_2 + ) = 2^n, so even sum = 2^{n-1}
Pause — copy the $x=-1$ result into your book: $\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \cdots = 0$ (for $n \ge 1$); adding $x=1$ and $x=-1$ results: $2(\binom{n}{0} + \binom{n}{2} + \cdots) = 2^n$, so even-index sum $= 2^{n-1}$.
Did you get this? True or false: $\,^5C_0 - \,^5C_1 + \,^5C_2 - \,^5C_3 + \,^5C_4 - \,^5C_5 = 0$.
Worked examples · 3 in a row, reveal as you go
Evaluate $\,^7C_0 + \,^7C_1 + \,^7C_2 + \,^7C_3 + \,^7C_4 + \,^7C_5 + \,^7C_6 + \,^7C_7$.
Find $\,^8C_0 - \,^8C_1 + \,^8C_2 - \,^8C_3 + \,^8C_4 - \,^8C_5 + \,^8C_6 - \,^8C_7 + \,^8C_8$.
Find $\,^4C_0 + 2\cdot\,^4C_1 + 4\cdot\,^4C_2 + 8\cdot\,^4C_3 + 16\cdot\,^4C_4$.
Fill the gap: $\,^5C_0 + 3\cdot\,^5C_1 + 9\cdot\,^5C_2 + 27\cdot\,^5C_3 + 81\cdot\,^5C_4 + 243\cdot\,^5C_5 = (1+3)^5 = $ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: to evaluate $\,^n C_0 + 2\,^nC_1 + 4\,^nC_2 + \cdots + 2^n\,^nC_n$, the correct substitution is $x = 2$.
Activities · practice with the ideas
Evaluate $\,^{10}C_0 + \,^{10}C_1 + \cdots + \,^{10}C_{10}$.
Find $\,^9C_0 - \,^9C_1 + \,^9C_2 - \cdots - \,^9C_9$.
Evaluate $\,^5C_0 + 2\cdot\,^5C_1 + 4\cdot\,^5C_2 + 8\cdot\,^5C_3 + 16\cdot\,^5C_4 + 32\cdot\,^5C_5$.
Show that $\,^nC_0 + \,^nC_2 + \,^nC_4 + \cdots = 2^{n-1}$ by adding the $x=1$ and $x=-1$ results.
Evaluate $\,^6C_1 + 2\cdot\,^6C_2 + 3\cdot\,^6C_3 + 4\cdot\,^6C_4 + 5\cdot\,^6C_5 + 6\cdot\,^6C_6$. (Hint: differentiate $(1+x)^6$, then substitute $x=1$.)
Odd one out: Which of the following sums does NOT equal $2^n$ for some integer $n$?
Earlier you were asked: what value of $x$ gives the plain sum, and what gives the alternating sum?
$x = 1$ gives the plain sum: $(1+1)^n = 2^n = \,^nC_0 + \,^nC_1 + \cdots + \,^nC_n$.
$x = -1$ gives the alternating sum: $(1-1)^n = 0 = \,^nC_0 - \,^nC_1 + \,^nC_2 - \cdots$.
And for the opening challenge: $\,^{100}C_0 + \,^{100}C_1 + \cdots + \,^{100}C_{100} = (1+1)^{100} = 2^{100}$. That's a number with 31 digits — found in a single step.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Evaluate $\,^7C_0 + \,^7C_1 + \,^7C_2 + \cdots + \,^7C_7$ using the binomial theorem. (2 marks)
Q2. Find $\,^4C_0 + 2\cdot\,^4C_1 + 4\cdot\,^4C_2 + 8\cdot\,^4C_3 + 16\cdot\,^4C_4$. (2 marks)
Q3. (a) Prove that $\sum_{r=0}^{n} \,^nC_r = 2^n$ using the binomial theorem. (b) Hence show that $\,^nC_0 + \,^nC_2 + \,^nC_4 + \cdots = 2^{n-1}$. (c) Explain the combinatorial meaning of both results. (3 marks)
Comprehensive answers (click to reveal)
Activities: 1. $2^{10}=1024$ · 2. $(1-1)^9=0$ · 3. $(1+2)^5=3^5=243$ · 4. Adding: $2(\,^nC_0+\,^nC_2+\cdots)=2^n$, so sum $=2^{n-1}$ · 5. Differentiating $(1+x)^6 = \sum \,^6C_r x^r$ gives $6(1+x)^5 = \sum r\,^6C_r x^{r-1}$; set $x=1$: $6\cdot 2^5 = 192$.
Q1 (2 marks): Set $x=1$ in $(1+x)^7 = \sum_{r=0}^7 \,^7C_r x^r$ [1]. LHS $= 2^7 = 128$ [1].
Q2 (2 marks): Rewrite: $\sum_{r=0}^4 \,^4C_r \cdot 2^r$ with $x=2$ [1]. $(1+2)^4 = 3^4 = 81$ [1].
Q3 (3 marks): (a) Set $x=1$: $(1+1)^n = \sum \,^nC_r \Rightarrow 2^n = \sum \,^nC_r$ [1]. (b) Set $x=-1$: $0=\,^nC_0-\,^nC_1+\cdots$. Add: $2^n = 2(\,^nC_0+\,^nC_2+\cdots)$, giving $2^{n-1}$ [1]. (c) $2^n$ = number of subsets of an $n$-set; $2^{n-1}$ = subsets of even size (equals subsets of odd size) [1].
Five timed questions drawn from the Module 4 bank. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering Combinatorics questions. Lighter alternative to the boss.
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