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Manipal Entrance Test 2016 Math Paper
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© examsnet.com
Question : 7
Total: 70
If
t
u
=
n
Σ
r
=
0
1
(
n
C
r
)
k
and
S
n
=
n
Σ
r
=
0
r
(
n
C
r
)
k
, where
k
∈
Z
+
,
then
cos
−
1
(
S
n
n
t
n
)
is
π
6
π
3
π
4
π
2
Validate
Solution:
S
n
=
0
(
n
C
0
)
k
+
1
(
n
C
1
)
k
+
2
(
n
C
2
)
k
+
.
.
.
.
+
n
(
n
C
n
)
k
⇒
S
n
=
n
(
n
C
0
)
k
+
n
−
1
(
n
C
1
)
k
+
n
−
2
(
n
C
2
)
k
+
.
.
.
.
+
0
(
n
C
0
)
k
⇒
2
S
n
=
n
t
n
⇒
S
n
n
t
n
=
1
2
Now,
cos
−
1
(
S
n
n
t
n
)
=
cos
−
1
(
1
2
)
=
π
3
© examsnet.com
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