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IIT JEE Advanced 2008 Paper 1
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© examsnet.com
Question : 56
Total: 69
Let
S
n
=
n
Σ
k
=
1
n
n
2
+
k
n
+
k
2
and
T
n
=
n
−
1
Σ
k
=
0
n
n
2
+
k
n
+
k
2
fo n = 1 , 2 , 3 , ... , then ,
S
n
<
π
3
√
3
S
n
>
π
3
√
3
T
n
<
π
3
√
3
T
n
>
π
3
√
3
Validate
Solution:
We have
s
n
=
n
Σ
k
=
1
n
n
2
+
n
k
+
k
2
⇒
s
n
<
s
∞
s
n
>
lim
n
→
∞
n
Σ
k
=
1
n
n
2
+
n
k
+
k
2
=
lim
n
→
∞
n
Σ
k
=
1
1
n
1
+
k
2
+
(
k
n
)
2
=
1
∫
0
d
x
1
+
x
+
x
2
=
π
3
√
3
⇒
s
n
<
π
3
√
3
h
n
Σ
k
=
1
f (kh) <
1
∫
0
f (x) dx <
h
n
−
1
Σ
k
=
0
f (kh)
⇒
T
n
>
π
3
√
3
© examsnet.com
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