Binomial theorem — statement, proof by induction: Exercises

Before we start, one reminder of the single tool everything uses — the general term: $T_{k + 1} = (k n) a^{n - k} b^{k}, (k n) = \frac{n !}{k ! ( n - k )!} .$ Read it as: "choose $k$ of the $n$ brackets to give $b$ ; the first letter falls from power $n$ ; the second letter rises from power $0$ ." The powers of the two letters always add up to $n$ — keep that in your back pocket for every single problem.

Figure — Binomial theorem — statement, proof by induction

Level 1 — Recognition

L1.1 — Count the terms

How many terms appear in the expansion of $(x + y)^{12}$ ?

Recall Solution

WHAT we use: the term count of a binomial expansion. WHY: the exponent of $b$ runs over every whole number from $0$ up to $n$ , so there is one term for each of those values. The values $k = 0, 1, 2, \dots, 12$ give $12 - 0 + 1 = 13$ terms. $13 terms.$

L1.2 — Read off a coefficient from Pascal's Triangle

Using Pascal's Triangle, write the full expansion of $(a + b)^{5}$ .

Recall Solution

Row $n = 5$ of the triangle (start counting rows at $0$ ) is $1, 5, 10, 10, 5, 1.$ These are $(0 5), (1 5), (2 5), (3 5), (4 5), (5 5)$ . Pair each with $a^{5 - k} b^{k}$ (first letter falls, second rises): $(a + b)^{5} = a^{5} + 5 a^{4} b + 10 a^{3} b^{2} + 10 a^{2} b^{3} + 5 a b^{4} + b^{5} .$ Check: put $a = b = 1$ : RHS $= 1 + 5 + 10 + 10 + 5 + 1 = 32 = 2^{5}$ . ✓

L1.3 — Which term is which?

In $(a + b)^{10}$ , what is the value of $k$ that produces the 7th term $T_{7}$ ?

Recall Solution

WHY the shift: the general term is written $T_{k + 1}$ because $k$ starts at $0$ . So the term index is one more than $k$ . Set $k + 1 = 7 \Rightarrow k = 6$ . That term is $T_{7} = (6 10) a^{4} b^{6}$ . $k = 6.$

Level 2 — Application

L2.1 — Full expansion with a constant

Expand $(2 x + 3)^{4}$ completely.

Recall Solution

WHAT: apply $T_{k + 1} = (k 4) (2 x)^{4 - k} (3)^{k}$ . Notice $a = 2 x$ (with its coefficient!) and $b = 3$ .

$k = 0 : (0 4) (2 x)^{4} = 16 x^{4}$
$k = 1 : (1 4) (2 x)^{3} (3) = 4 \cdot 8 x^{3} \cdot 3 = 96 x^{3}$
$k = 2 : (2 4) (2 x)^{2} (9) = 6 \cdot 4 x^{2} \cdot 9 = 216 x^{2}$
$k = 3 : (3 4) (2 x) (27) = 4 \cdot 2 x \cdot 27 = 216 x$
$k = 4 : (4 4) (81) = 81$ $(2 x + 3)^{4} = 16 x^{4} + 96 x^{3} + 216 x^{2} + 216 x + 81.$ Check: put $x = 1$ : LHS $= (2 + 3)^{4} = 625$ ; RHS $= 16 + 96 + 216 + 216 + 81 = 625$ . ✓

L2.2 — A negative second term

Expand $(x - 2)^{5}$ .

Recall Solution

WHY the sign matters: the second letter is the whole thing $(- 2)$ , so each term carries $(- 2)^{k}$ , which flips sign on odd $k$ . Coefficients from row 5: $1, 5, 10, 10, 5, 1$ . $T_{k + 1} = (k 5) x^{5 - k} (- 2)^{k} .$

$k = 0 : x^{5}$
$k = 1 : 5 x^{4} (- 2) = - 10 x^{4}$
$k = 2 : 10 x^{3} (4) = 40 x^{3}$
$k = 3 : 10 x^{2} (- 8) = - 80 x^{2}$
$k = 4 : 5 x (16) = 80 x$
$k = 5 : (- 32) = - 32$ $(x - 2)^{5} = x^{5} - 10 x^{4} + 40 x^{3} - 80 x^{2} + 80 x - 32.$ Check: put $x = 1$ : LHS $= (1 - 2)^{5} = - 1$ ; RHS $= 1 - 10 + 40 - 80 + 80 - 32 = - 1$ . ✓

L2.3 — A single coefficient without expanding

Find the coefficient of $x^{5}$ in $(1 + x)^{9}$ .

Recall Solution

Using the $(1 + x)^{n}$ form (parent note, Example 3): $(1 + x)^{9} = \sum_{k} (k 9) x^{k}$ . The coefficient of $x^{5}$ is just $(5 9)$ . $(5 9) = \frac{9 !}{5 ! 4 !} = \frac{9 \cdot 8 \cdot 7 \cdot 6}{4 \cdot 3 \cdot 2 \cdot 1} = 126.$ $126.$

Level 3 — Analysis

L3.1 — Term free of $x$ (the constant term)

Find the term independent of $x$ in $(x^{2} + \frac{1}{x})^{9}$ .

Recall Solution

WHAT: write the general term and track the total power of $x$ . $T_{k + 1} = (k 9) (x^{2})^{9 - k} (\frac{1}{x})^{k} = (k 9) x^{2 (9 - k)} x^{- k} = (k 9) x^{18 - 3 k} .$ WHY set the power to zero: "independent of $x$ " means $x^{0}$ , i.e. the exponent is $0$ . $18 - 3 k = 0 \Rightarrow k = 6.$ $T_{7} = (6 9) = \frac{9 !}{6 ! 3 !} = 84.$ $84.$

L3.2 — A coefficient with a sign and a power

Find the coefficient of $x^{3}$ in $(x - \frac{2}{x})^{7}$ .

Recall Solution

$T_{k + 1} = (k 7) x^{7 - k} (- \frac{2}{x})^{k} = (k 7) (- 2)^{k} x^{7 - k - k} = (k 7) (- 2)^{k} x^{7 - 2 k} .$ Set the power equal to $3$ : $7 - 2 k = 3 \Rightarrow k = 2$ . $(2 7) (- 2)^{2} = 21 \cdot 4 = 84.$ $84.$

L3.3 — Middle term

Find the middle term of $(3 - x)^{6}$ .

Recall Solution

WHY there is exactly one middle term: $n = 6$ is even, so there are $n + 1 = 7$ terms; the single middle one is term number $\frac{n}{2} + 1 = 4$ , i.e. $k = 3$ . (See General term & middle term of a binomial expansion.) $T_{4} = (3 6) (3)^{6 - 3} (- x)^{3} = 20 \cdot 27 \cdot (- x^{3}) = - 540 x^{3} .$ $- 540 x^{3} .$

Level 4 — Synthesis

L4.1 — Two conditions at once

In $(2 x^{2} - \frac{1}{x})^{9}$ , find the coefficient of $x^{6}$ (parent Example 2) and the term independent of $x$ .

Recall Solution

$T_{k + 1} = (k 9) (2 x^{2})^{9 - k} (- \frac{1}{x})^{k} = (k 9) 2^{9 - k} (- 1)^{k} x^{18 - 2 k - k} = (k 9) 2^{9 - k} (- 1)^{k} x^{18 - 3 k} .$ Coefficient of $x^{6}$ : $18 - 3 k = 6 \Rightarrow k = 4$ . $(4 9) 2^{5} (- 1)^{4} = 126 \cdot 32 = 4032.$ Independent of $x$ : $18 - 3 k = 0 \Rightarrow k = 6$ . $(6 9) 2^{3} (- 1)^{6} = 84 \cdot 8 = 672.$ $Coeff of x^{6} = 4032, constant term = 672.$

L4.2 — Evaluating a numeric sum via the theorem

Without a calculator, evaluate $k = 0 \sum 8 (k 8) 3^{k}$ .

Recall Solution

WHY the theorem helps: the sum looks exactly like $(1 + x)^{n} = \sum (k n) x^{k}$ with $n = 8$ and $x = 3$ — it is a binomial expansion in disguise. $\sum_{k = 0}^{8} (k 8) 1^{8 - k} 3^{k} = (1 + 3)^{8} = 4^{8} = 65536.$ $65536.$

L4.3 — Ratio of consecutive coefficients

In $(1 + x)^{n}$ , the coefficients of the $5$ th and $6$ th terms are equal. Find $n$ .

Recall Solution

WHAT: 5th term has $k = 4$ , 6th term has $k = 5$ . Coefficients are $(4 n)$ and $(5 n)$ . Set them equal: $(4 n) = (5 n) .$ WHY this gives a clean equation: $(k n) = (n - k n)$ , so two binomial coefficients from the same row are equal only when their lower indices are equal or sum to $n$ . Here $4 \neq = 5$ , so we need $4 + 5 = n$ . $n = 9.$ Check the direct way: $\frac{( 5 n )}{( 4 n )} = \frac{n - 4}{5} = 1 \Rightarrow n - 4 = 5 \Rightarrow n = 9$ . ✓

Level 5 — Mastery

L5.1 — Approximate a power

Use the first three terms of the binomial expansion to estimate $(1.02)^{10}$ , and compare with the exact value.

Recall Solution

WHY expand around $1$ : write $1.02 = 1 + 0.02$ so that $x = 0.02$ is small — later terms shrink fast, so a few terms already approximate well (this idea generalises in Binomial series for any index). $(1 + x)^{10} = 1 + (1 10) x + (2 10) x^{2} + \dots = 1 + 10 x + 45 x^{2} + \dots$ With $x = 0.02$ : $1 + 10 (0.02) + 45 (0.0004) = 1 + 0.2 + 0.018 = 1.218.$ $(1.02)^{10} \approx 1.218.$ Check: the exact value is $1.0 2^{10} = 1.21899 \dots$ , so three terms are already accurate to three decimals. ✓

L5.2 — Prove an identity using the theorem

Show that $k = 0 \sum n k (k n) = n 2^{n - 1}$ for $n \geq 1$ , then verify it at $n = 4$ .

Recall Solution

WHAT tool and WHY: we need to turn $k (k n)$ into something summable. Use the absorption identity $k (k n) = k \cdot \frac{n !}{k ! ( n - k )!} = \frac{n !}{( k - 1 )! ( n - k )!} = n \cdot \frac{( n - 1 )!}{( k - 1 )! ( n - k )!} = n (k - 1 n - 1) .$ (The factor $k$ cancels one factor of the $k!$ in the denominator — that is why absorption works.) Now sum, and re-index with $j = k - 1$ (limits shift $0 \to n$ becomes $- 1 \to n - 1$ , but the $k = 0$ term is zero so start at $k = 1$ , i.e. $j = 0$ ): $\sum_{k = 1}^{n} n (k - 1 n - 1) = n \sum_{j = 0}^{n - 1} (j n - 1) = n \cdot 2^{n - 1},$ using $\sum_{j} (j n - 1) = 2^{n - 1}$ (put $a = b = 1$ in the theorem for exponent $n - 1$ ). $■$ Verify at $n = 4$ : LHS $= 0 \cdot 1 + 1 \cdot 4 + 2 \cdot 6 + 3 \cdot 4 + 4 \cdot 1 = 0 + 4 + 12 + 12 + 4 = 32$ . RHS $= 4 \cdot 2^{3} = 32$ . ✓

L5.3 — Reconstruct one line of the induction proof

In the inductive step we had $(a + b)^{n + 1} = \sum_{k = 0}^{n} (k n) a^{n - k + 1} b^{k} + \sum_{k = 0}^{n} (k n) a^{n - k} b^{k + 1} .$ Take $n = 3$ and $k$ -slot power $a^{2} b^{2}$ (i.e. the $a^{(n + 1) - k} b^{k}$ term with $k = 2$ ). Show numerically that the two sums contribute $(2 3)$ and $(1 3)$ , and that they add to $(2 4) = 6$ .

Recall Solution

WHERE the two pieces come from:

The first sum has term $(k n) a^{n - k + 1} b^{k}$ . For $a^{2} b^{2}$ with $n = 3$ we need $k = 2$ : that gives $(2 3) a^{2} b^{2} = 3 a^{2} b^{2}$ .
The second sum has term $(k n) a^{n - k} b^{k + 1}$ . For $a^{2} b^{2}$ we need $k + 1 = 2$ , i.e. $k = 1$ : that gives $(1 3) a^{2} b^{2} = 3 a^{2} b^{2}$ . Adding, the coefficient of $a^{2} b^{2}$ in $(a + b)^{4}$ is $(2 3) + (1 3) = 3 + 3 = 6 = (2 4),$ which is exactly Pascal's rule $(k n) + (k - 1 n) = (k n + 1)$ at $n = 3, k = 2$ . This is the engine of the whole proof, seen on one term. $(2 3) + (1 3) = 6 = (2 4) .$

Recall One-line self-test summary

Every problem here reduces to one move: write $T_{k + 1} = (k n) a^{n - k} b^{k}$ , plug in the actual $a$ and $b$ (with signs and coefficients), collect the power of $x$ , and set it to the target. The single move that solves 90% of binomial problems? ::: Write the general term, collect the power of $x$ , solve for $k$ .

Connections

Parent: Binomial Theorem — the statement and induction proof these exercises drill.
General term & middle term of a binomial expansion — L3.1–L3.3 are direct applications.
Combinations — nCr and Factorials — evaluating every $(k n)$ .
Pascal's Triangle — L1.2 and the Pascal-rule check in L5.3.
Mathematical Induction — the framework behind L5.3.
Binomial series for any index — the approximation idea in L5.1 generalises here.