Standard +0.8 This is a standard proof by induction with a summation involving exponentials. While it requires careful algebraic manipulation in the inductive step (combining fractions with powers of 5 and simplifying to reach the required form), it follows the routine induction framework without requiring novel insight. The algebra is moderately demanding but manageable for Further Maths students, placing it slightly above average difficulty.
Formula is true for n(cid:32)1since (cid:32) 1 and
51 5
5(cid:16)4(cid:117)1
(cid:32) 1
5 5
Assume that the formula is true for n(cid:32)k
k(cid:14)15(cid:16)4r k 5(cid:16)4(k(cid:14)1)
Then (cid:166) (cid:32) (cid:14)
5 5k 5k(cid:14)1
r(cid:32)1
5k(cid:14)5(cid:16)4(k(cid:14)1)
(cid:32)
5k(cid:14)1
k(cid:14)1
(cid:32)
5k(cid:14)1
So if true for n(cid:32)k then true also for n(cid:32)k(cid:14)1
Answer
Marks
So true generally
B1
M1
M1
A1
E1
Answer
Marks
[5]
2.5
3.1a
2.1
2.2a
i
Answer
Marks
2.4
Must include an arithmetic justification
n
Add extra term
e
Manipulate
m
Needs conclusion
9
n
e
m
i
Answer
Marks
10(i)(a)
c
e
p
S
10(i)(b)
10(i)(c)
10(i)(d)
10(i)(e)
n
e
m
i
c
e
Question 9:
9 | 1
Formula is true for n(cid:32)1since (cid:32) 1 and
51 5
5(cid:16)4(cid:117)1
(cid:32) 1
5 5
Assume that the formula is true for n(cid:32)k
k(cid:14)15(cid:16)4r k 5(cid:16)4(k(cid:14)1)
Then (cid:166) (cid:32) (cid:14)
5 5k 5k(cid:14)1
r(cid:32)1
5k(cid:14)5(cid:16)4(k(cid:14)1)
(cid:32)
5k(cid:14)1
k(cid:14)1
(cid:32)
5k(cid:14)1
So if true for n(cid:32)k then true also for n(cid:32)k(cid:14)1
So true generally | B1
M1
M1
A1
E1
[5] | 2.5
3.1a
2.1
2.2a
i
2.4 | Must include an arithmetic justification
n
Add extra term
e
Manipulate
m
Needs conclusion
9
n
e
m
i
10(i)(a) | c
e
p
S
10(i)(b)
10(i)(c)
10(i)(d)
10(i)(e)
n
e
m
i
c
e
9 Prove by induction that, for all positive integers $n$,
$$\sum _ { r = 1 } ^ { n } \frac { 5 - 4 r } { 5 ^ { r } } = \frac { n } { 5 ^ { n } }$$
\hfill \mbox{\textit{OCR Further Pure Core 1 Q9 [5]}}