SME
Grade 12 Math · Problem type
A geometric sum
Use this for “Prove 1 + 2 + 4 + … + 2ⁿ⁻¹ = 2ⁿ − 1 by induction” (r = 2), or any ratio r.
Chart
Terms: 1, 2, 4, 8, 16 (multiply by r = 2 each time).
a₅ = 1 × 2⁴ = 16
S₅
=
1
×
1 −
2
5
1 − 2
=
31
This is an example. Type your own values to start a new problem.
1
+
r
+
…
+
r
−
1
=
r
−
1
−
1
=
S
a
Next term
worked out (not typed)
T
S plus the next term
worked out (not typed)
F
The formula at n + 1
worked out (not typed)
Formulas
S
=
r
n
− 1
r
− 1
31
=
2
5
− 1
2 − 1
a
=
r
n
32
=
2
5
T
=
S
+
a
63 = 31 + 32
F
=
r
n
+ 1
− 1
r
− 1
63
=
2
5 + 1
− 1
2 − 1
Assumptions
•
Base case: check the formula at n = 1.
•
Step: assume it holds at n = k; adding term k + 1 must give the formula at k + 1.
•
Here: (rᵏ − 1)/(r − 1) + rᵏ = (rᵏ⁺¹ − 1)/(r − 1), since rᵏ(r − 1) = rᵏ⁺¹ − rᵏ.
Step-by-step
WE KNOW
r = 2, n = 5
FIND
sum of the first n terms (S), next term (a), S plus the next term (T), the formula at n + 1 (F)
Step 1 · Find sum of the first n terms (S)
Use S = (rⁿ − 1) ÷ (r − 1)
The
formula
at
n
:
1
+
r
+
…
+
rⁿ⁻¹
=
(rⁿ
−
1)/(
r
−
1).
S
=
2
5
− 1
2 − 1
S
=
32
−
1
S
=
31
Step 2 · Find next term (a)
Use a = rⁿ
The
next
term,
number
n
+
1:
r
to
the
power
n
.
a
=
2
5
a
=
32
Step 3 · Find S plus the next term (T)
Use T = S + a
If
the
formula
holds
at
n
,
the
sum
through
term
n
+
1
is
S
plus
the
next
term.
T
=
31
+
32
T
=
63
Step 4 · Find the formula at n + 1 (F)
Use F = (rⁿ⁺¹ − 1) ÷ (r − 1)
The
formula
with
n
+
1
in
place
of
n
.
F
=
2
5 + 1
− 1
2 − 1
F
=
2
6
− 1
2 − 1
F
=
64
−
1
F
=
63
→
T
=
F
=
63:
adding
the
next
term
gives
the
formula
at
n
+
1.
With
k
for
n
:
(rᵏ
−
1)/(
r
−
1)
+
rᵏ
=
(rᵏ⁺¹
−
1)/(
r
−
1)
Check
Put the numbers back in. Both sides should match.
31
=
2
5
− 1
2 − 1
✓
32
=
2
5
✓
63 = 31 + 32 ✓
63
=
2
5 + 1
− 1
2 − 1
✓
Topic check
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