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2010
P1 Q2
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Maclaurin
10 P1 Q2
2010 H2 Mathematics Paper 1 Question 2
Maclaurin Series
Answers
(i)
1
+
3
x
+
5
2
x
2
{1 + 3 x + \frac{5}{2} x^2}
1
+
3
x
+
2
5
x
2
(ii)
n
=
9
4
{n=\frac{9}{4}}
n
=
4
9
Full solutions
(i)
e
x
(
1
+
sin
2
x
)
=
(
1
+
x
+
1
2
x
2
+
…
)
(
1
+
2
x
+
…
)
=
1
+
3
x
+
5
2
x
2
+
…
■
\begin{align*} & \mathrm{e}^x (1 + \sin 2x) \\ &= \left( 1 + x + \frac{1}{2} x^2 + \ldots \right) \left( 1 + 2 x + \ldots \right) \\ &= 1 + 3 x + \frac{5}{2} x^2 + \ldots \; \blacksquare \end{align*}
e
x
(
1
+
sin
2
x
)
=
(
1
+
x
+
2
1
x
2
+
…
)
(
1
+
2
x
+
…
)
=
1
+
3
x
+
2
5
x
2
+
…
■
(ii)
(
1
+
4
3
x
)
n
=
1
+
4
3
n
x
+
n
(
n
−
1
)
2
16
9
x
2
+
…
\begin{align*} & \left( 1 + \frac{4}{3}x \right )^n \\ & = 1 + \frac{4}{3}nx + \frac{n(n-1)}{2} \frac{16}{9}x^2 + \ldots \\ \end{align*}
(
1
+
3
4
x
)
n
=
1
+
3
4
n
x
+
2
n
(
n
−
1
)
9
16
x
2
+
…
Since the first two terms are equal,
4
3
n
=
3
n
=
9
4
■
\begin{align*} \frac{4}{3} n &= 3 \\ n &= \frac{9}{4} \; \blacksquare \end{align*}
3
4
n
n
=
3
=
4
9
■
Third term of
(
1
+
4
3
x
)
n
=
9
4
(
9
4
−
1
)
2
16
9
x
2
=
5
2
x
2
\begin{align*} & \textrm{Third term of } \left( {\textstyle 1 + \frac{4}{3}x} \right )^n \\ & = \frac{\frac{9}{4}\left(\frac{9}{4}-1\right)}{2} \frac{16}{9}x^2 \\ &= \frac{5}{2}x^2 \end{align*}
Third term of
(
1
+
3
4
x
)
n
=
2
4
9
(
4
9
−
1
)
9
16
x
2
=
2
5
x
2
Hence the third terms in each of these series are equal
■
{\blacksquare}
■
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