STEP I, 2003
2
Section A:
Pure Mathematics
1
It is given that
n
∑
r
=

1
r
2
can be written in the form
pn
3
+
qn
2
+
rn
+
s
, where
p
,
q
,
r
and
s
are numbers. By setting
n
=

1, 0, 1 and 2, obtain four equations that must be satisﬁed by
p
,
q
,
r
and
s
and hence show that
n
∑
r
=0
r
2
=
1
6
n
(
n
+ 1)(2
n
+ 1)
.
Given that
n
∑
r
=

2
r
3
can be written in the form
an
4
+
bn
3
+
cn
2
+
dn
+
e
, show similarly that
n
∑
r
=0
r
3
=
1
4
n
2
(
n
+ 1)
2
.
2
The ﬁrst question on an examination paper is:
Solve for
x
the equation
1
x
=
1
a
+
1
b
.
where (in the question)
a
and
b
are given nonzero real numbers. One candidate writes
x
=
a
+
b
as the solution. Show that there are no values of
a
and
b
for which this will give the correct
answer.
The next question on the examination paper is:
Solve for
x
the equation
1
x
=
1
a
+
1
b
+
1
c
.
where (in the question)
a
,
b
and
c
are given nonzero numbers. The candidate uses the same
technique, giving the answer as
x
=
a
+
b
+
c .
Show that the candidate’s answer will be correct
if and only if
a
,
b
and
c
satisfy at least one of the equations
a
+
b
= 0 ,
b
+
c
= 0 or
c
+
a
= 0 .
(i)
3
Show that 2 sin(
1
2
θ
) = sin
θ
if and only if sin(
1
2
θ
) = 0 .
(ii)
Solve the equation 2 tan(
1
2
θ
) = tan
θ
.
(iii)
Show that 2 cos(
1
2
θ
) = cos
θ
if and only if
θ
= (4
n
+ 2)
π
±
2
φ
where
φ
is deﬁned by
cos
φ
=
1
2
(
√
3

1) , 0
6
φ
6
π/
2, and
n
is any integer.
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3
4
Solve the inequality
sin
θ
+ 1
cos
θ
6
1
where 0
6
θ <
2
π
and cos
θ
6
= 0 .
(i)
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 Spring '12
 rotar
 Math, Equations, Numerical digit, binomial expansion, nonzero real numbers, penalty shootout

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