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2010
P1 Q3
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AP/GP
10 P1 Q3
2010 H2 Mathematics Paper 1 Question 3
Arithmetic and Geometric Progressions (APs, GPs)
Answers
(i)
u
n
=
4
n
−
2
+
c
{u_n = 4n-2+c}
u
n
=
4
n
−
2
+
c
(ii)
Out of syllabus
Full solutions
(i)
u
n
=
S
n
−
S
n
−
1
=
n
(
2
n
+
c
)
−
(
n
−
1
)
(
2
(
n
−
1
)
+
c
)
=
2
n
2
+
c
n
−
(
2
(
n
2
−
2
n
+
1
)
+
c
(
n
−
1
)
)
=
2
n
2
+
c
n
−
(
2
n
2
−
4
n
+
2
+
c
n
−
c
)
=
4
n
−
2
+
c
■
\begin{align*} u_n &= S_{n} - S_{n-1} \\ &= n(2n+c) - (n-1)\Big(2(n-1)+c\Big) \\ &= 2n^2 + cn - (2(n^2-2n+1)+c(n-1)) \\ &= 2n^2 + cn - (2n^2 - 4n + 2 + cn - c) \\ &= 4n - 2 + c \; \blacksquare \end{align*}
u
n
=
S
n
−
S
n
−
1
=
n
(
2
n
+
c
)
−
(
n
−
1
)
(
2
(
n
−
1
)
+
c
)
=
2
n
2
+
c
n
−
(
2
(
n
2
−
2
n
+
1
)
+
c
(
n
−
1
))
=
2
n
2
+
c
n
−
(
2
n
2
−
4
n
+
2
+
c
n
−
c
)
=
4
n
−
2
+
c
■
(ii)
Out of syllabus
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