SME
Grade 11 Math · Algebra 2
The binomial theorem and Pascal's triangle
Use this for “Find the coefficient of x³ in (2x − 3)⁵.”
Diagram
Row 5: 1, 5, 10, 10, 5, 1; they add to 2⁵ = 32.
Entry 3 of row 5: C(5, 3) = 10.
The two above add to it: C(4, 2) + C(4, 3) = 6 + 4 = 10.
(A + B)⁵ = A⁵ + 5A⁴B + 10A³B² + 10A²B³ + 5AB⁴ + B⁵
This is an example. Type your own values to start a new problem.
(
x
+
(
)
)
C(
n
,
k
)
×
a
×
(
)
−
=
T
The second line is the coefficient of xᵏ: k of the n brackets give ax, the rest give b.
C
Entry of Pascal’s triangle, C(n, k)
worked out (not typed)
Formulas
C
=
C
(
n
,
k
)
10
=
C
(5,
3)
T
=
C
×
a
k
×
b
n
−
k
720
=
10
×
2
3
×
(−3)
5 − 3
Assumptions
•
Write (ax + b)ⁿ as (A + B)ⁿ with A = ax and B = b: row n of Pascal’s triangle gives the coefficients.
•
In every term the powers of A and B add up to n: the xᵏ term is C(n, k)Aᵏ Bⁿ⁻ᵏ.
•
When exactly one of a and b is negative, the terms alternate in sign.
Step-by-step
WE KNOW
a = 2, b = −3, n = 5, k = 3
FIND
entry of Pascal’s triangle, C(n, k) (C), coefficient of xᵏ (T)
Step 1 · Find entry of Pascal’s triangle, C(n, k): C
Use C = C(n, k)
Entry
k
of
row
n
of
Pascal’s
triangle:
the
ways
to
pick
k
of
the
n
factors
for
ax.
C
=
C
(5,
3)
C
=
10
Step 2 · Find coefficient of xᵏ (T)
Use T = C × aᵏ × bⁿ⁻ᵏ
k
factors
give
ax
and
the
other
n
−
k
give
b
.
T
=
10
×
2
3
×
(−3)
5 − 3
T
=
10
×
2
3
×
(−3)
2
T
=
10
×
8
×
9
T
=
720
Check
Put the numbers back in. Both sides should match.
10
=
C
(5,
3)
✓
720
=
10
×
2
3
×
(−3)
5 − 3
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.