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.
Today's challenge, Without a calculator, find $\,{}^{100}C_0 + \,{}^{100}C_1 + \,{}^{100}C_2 + \cdots + \,{}^{100}C_{100}$. There are 101 terms to add. Is there a shortcut? By the end of this lesson you'll evaluate it in under 3 seconds, and explain exactly why it works.
0/5QUESTS
1
You’re here
Recall, your gut answer first
01
Recall, your gut answer first
+5 XP warm-up
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.
auto-saved
2
You’re here
Orientation and key terms
02
The two substitutions
+5 XP to read
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.
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).
Why $2^n$ counts subsets. For each of the $n$ elements in a set, you make a binary choice: include it or don't. That gives $2 \times 2 \times \cdots \times 2 = 2^n$ total subsets. And $\,{}^nC_r$ counts exactly the subsets of size $r$. Summing over all $r$ recovers $2^n$, the substitution proof mirrors this counting argument.
\,^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 $= 2^{n-1}$.
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 $= 2^{n-1}$.
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: $\,{}^5C_0 - \,{}^5C_1 + \,{}^5C_2 - \,{}^5C_3 + \,{}^5C_4 - \,{}^5C_5 = 0$.
Express each multiplier as a power of 2: $1=2^0$, $2=2^1$, $4=2^2$, $8=2^3$, $16=2^4$.
2
This is $\sum_{r=0}^{4} \,{}^4C_r \cdot 2^r$, exactly the binomial theorem with $x=2$.
Match to $(1+x)^4$ with $x=2$: each term $\,{}^4C_r x^r = \,{}^4C_r \cdot 2^r$. ✓
3
$(1+2)^4 = 3^4 = \mathbf{81}$.
Substitute $x=2$: the LHS is $(1+2)^4 = 81$. One calculation replaces five additions.
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 = $ .
5
You’re here
Misconceptions to fix
Misconceptions to fix · the 3 traps that cost marks
Trap 01
Not recognising the pattern
Students try to evaluate each $\,{}^nC_r$ individually and add. Look for alternating signs (use $x=-1$), powers-of-$a$ multipliers (use $x=a$), or no multiplier (use $x=1$) before touching the coefficients.
Trap 02
Wrong substitution value
The substitution must match what multiplies $\,{}^nC_r$. If the multipliers are $1, 2, 4, 8, \ldots = 2^r$, the substitution is $x=2$, not $x=2^n$. Carefully check that each term's coefficient matches $x^r$ for your chosen $x$.
Trap 03
Forgetting the expansion form
The standard form is $(1+x)^n$, not $(a+b)^n$. If a sum has coefficients $\,{}^nC_r a^r$, write it as $(1+a)^n$. If it has $\,{}^nC_r a^r b^{n-r}$, you need the full binomial theorem $(a+b)^n$.
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 $x = 2$.
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.
auto-saved
1
You’re here
Multiple choice
01
Multiple choice
+5 XP per correct · +25 XP all-correct
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
2
You’re here
Short answer
02
Short answer
ApplyBand 42 marks
Q1. Evaluate $\,{}^7C_0 + \,{}^7C_1 + \,{}^7C_2 + \cdots + \,{}^7C_7$ using the binomial theorem. (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)
