Binomial theorem — statement, proof by induction: Question bank

The theorem, kept next to you the whole time: $(a + b)^{n} = \sum_{k = 0}^{n} (k n) a^{n - k} b^{k}, (k n) = \frac{n !}{k ! ( n - k )!} .$

True or false — justify

$(a + b)^{n}$ has $n$ terms. ::: False. It has $== n + 1 ==$ terms: $k$ runs from $0$ to $n$ inclusive, and counting both endpoints gives $n + 1$ values.
The exponents of $a$ and $b$ in every single term add up to $n$ . ::: True. A term is $a^{n - k} b^{k}$ , so the exponents sum to $(n - k) + k = n$ for every $k$ — this is the fingerprint of a genuine binomial term.
In $(a + b)^{n}$ the coefficients read the same forwards and backwards. ::: True. Since $(k n) = (n - k n)$ (choosing $k$ to include = choosing $n - k$ to exclude), the coefficient list is a palindrome, e.g. $1, 4, 6, 4, 1$ .
Every coefficient in the expanded $(a - b)^{n}$ is positive. ::: False. The magnitudes $(k n)$ are positive, but the signs alternate because of $(- 1)^{k}$ : e.g. $(a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}$ .
$(k n)$ is always a whole number. ::: True. It counts the number of ways to choose $k$ objects from $n$ — you cannot have a fractional number of choices, so despite the factorial fractions it is always an integer.
The binomial theorem as stated (finite sum) works for $n = \frac{1}{2}$ . ::: False. The finite-sum form needs a positive integer $n$ so the sum stops at $k = n$ . Non-integer indices give an infinite binomial series valid only for $∣ x ∣ < 1$ .
Setting $a = b = 1$ proves $\sum_{k = 0}^{n} (k n) = 2^{n}$ . ::: True. The theorem evaluated at $a = b = 1$ gives $2^{n}$ on the left and $\sum (k n)$ on the right — this also equals the total number of subsets of an $n$ -element set.
The theorem is only valid when $a$ and $b$ are positive numbers. ::: False. The proof uses only algebra (distribution and induction), so $a, b$ may be any real or complex numbers, including negatives and zero.

Spot the error

" $(a + b)^{3} = a^{3} + b^{3}$ because you cube each part." ::: The error is ignoring the cross terms. Expanding $(a + b) (a + b) (a + b)$ also lets you pick a mix of $a$ 's and $b$ 's, giving $a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ .
"In $(2 x - \frac{1}{x})^{5}$ the general term is $T_{k + 1} = (k 5) (2 x)^{5 - k} (\frac{1}{x})^{k}$ ." ::: The minus sign is missing. The second term is $- \frac{1}{x}$ , so it must enter with its sign: $T_{k + 1} = (k 5) (2 x)^{5 - k} (- \frac{1}{x})^{k}$ , carrying a $(- 1)^{k}$ .
"The 4th term of $(a + b)^{n}$ uses $k = 4$ in $(k n)$ ." ::: Off by one. Because $k$ starts at $0$ , the term index is $k + 1$ ; the 4th term has $k = 3$ , i.e. use $k = r - 1$ for the $r$ -th term.
"Pascal's rule is $(k n) + (k n) = (k n + 1)$ ." ::: The two terms must be neighbours, not identical: $(k - 1 n) + (k n) = (k n + 1)$ — an entry equals the sum of the two entries above it in Pascal's Triangle.
"During the induction step, after substituting $j = k + 1$ the limits stay $0$ to $n$ ." ::: The limits shift with the substitution. If $k$ ran $0 \to n$ , then $j = k + 1$ runs $1 \to n + 1$ ; forgetting this loses or duplicates the endpoint terms.
"The coefficient of $x^{2}$ in $(3 + x)^{4}$ is $(2 4) = 6$ ." ::: It forgot the powers of the other term. The $x^{2}$ term is $(2 4) 3^{2} x^{2} = 6 \cdot 9 x^{2} = 54 x^{2}$ ; the coefficient is $54$ , not $6$ .
" $(1 + x)^{n}$ has its largest coefficient at the first term." ::: The largest coefficient sits in the middle (or two middle) terms, since $(k n)$ grows then shrinks; the ends are the smallest ( $(0 n) = (n n) = 1$ ).

Why questions

Why is the coefficient of $a^{n - k} b^{k}$ exactly $(k n)$ ? ::: Because building that term means grabbing $b$ from exactly $k$ of the $n$ brackets, and the number of ways to choose which $k$ brackets is precisely $(k n)$ .
Why does the induction proof need Pascal's rule specifically? ::: After multiplying by $(a + b)$ and re-indexing, the two sums share the power $a^{(n + 1) - k} b^{k}$ with coefficient $(k n) + (k - 1 n)$ — Pascal's rule is exactly what collapses that into the single coefficient $(k n + 1)$ .
Why do we re-index the second sum with $j = k + 1$ instead of leaving it? ::: To align equal powers: the two sums originally carry different exponents on $b$ , and shifting the index makes both read $a^{(n + 1) - k} b^{k}$ so like terms can be added.
Why does $\sum_{k = 0}^{n} (- 1)^{k} (k n) = 0$ for $n \geq 1$ ? ::: It is the theorem at $a = 1, b = - 1$ , giving $(1 - 1)^{n} = 0^{n} = 0$ ; the positive and negative binomial coefficients exactly cancel in equal-sized batches.
Why do the powers of $a$ decrease while the powers of $b$ increase across the terms? ::: In each term the exponents must sum to $n$ , so as you grab $b$ from one more bracket (raising $b$ 's power by $1$ ) you grab $a$ from one fewer (lowering $a$ 's power by $1$ ).
Why can we not just say " $(a + b)^{n}$ is symmetric so $a^{3} + b^{3}$ is fine"? ::: Symmetry of the coefficients is real, but symmetry does not mean the cross terms vanish — the middle coefficients $(k n)$ for $0 < k < n$ are genuinely nonzero.

Edge cases

What is $(a + b)^{0}$ , and how many terms does it have? ::: It equals $1$ — a single term. With $n = 0$ the sum has $n + 1 = 1$ term, namely $(0 0) a^{0} b^{0} = 1$ , matching the convention $0! = 1$ (see Factorials).
What does the theorem give when $b = 0$ ? ::: Every term with $k \geq 1$ carries a factor $b^{k} = 0$ and vanishes, leaving only $k = 0$ : $(0 n) a^{n} = a^{n}$ , exactly as $(a + 0)^{n}$ should.
Is $(0 n)$ equal to $1$ , and why does that make sense? ::: Yes: $(0 n) = \frac{n !}{0 ! n !} = 1$ since $0! = 1$ ; there is exactly one way to choose nothing (the empty choice), so the first coefficient is always $1$ .
What happens to the general term when $k$ exceeds $n$ (say $k = n + 1$ )? ::: It does not exist in a positive-integer expansion; $(k n) = 0$ for $k > n$ because you cannot choose more brackets than there are, so the sum correctly stops at $k = n$ .
Does $(a - b)^{n}$ with an even $n$ have a symmetric sign pattern? ::: The magnitudes stay palindromic, but signs follow $(- 1)^{k}$ : for even $n$ the first and last terms are both positive, while for odd $n$ the last term is negative — always check parity of $k$ .
If $a = b$ , do any terms collapse or cancel? ::: Nothing cancels; instead every term becomes $(k n) a^{n}$ , and summing gives $2^{n} a^{n} = (2 a)^{n}$ , consistent with $(a + a)^{n} = (2 a)^{n}$ .

Recall One-line self-test before you leave

Cover and answer: How many terms, what is the sign law, and which lemma drives the proof? Terms $= n + 1$ ; the sign in $(a - b)^{n}$ is $(- 1)^{k}$ ; the proof is powered by Pascal's rule.

Connections

Parent: Binomial theorem — statement & proof
Pascal's Triangle — the palindrome and Pascal's-rule traps live here.
Combinations — nCr — why $(k n)$ is a whole number and why $(0 n) = 1$ .
Mathematical Induction — the re-indexing and limit-shift traps.
General term & middle term of a binomial expansion — the " $k$ -th vs $(k + 1)$ -th term" trap.
Binomial series for any index — the "does it work for $n = \frac{1}{2}$ ?" edge case.
Factorials — the $0! = 1$ convention behind the $n = 0$ edge case.