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Let
P
=
∑
n
=
1
11
n
(
1
2
1
+
n
+
2
2
2
+
n
+
3
2
3
+
n
+
⋯
+
11
2
11
+
n
)
P = \sum_{n=1}^{11} n \left( \frac{1^2}{1+n} + \frac{2^2}{2+n} + \frac{3^2}{3+n} + \dots + \frac{11^2}{11+n} \right)
P
=
∑
n
=
1
11
n
(
1
+
n
1
2
+
2
+
n
2
2
+
3
+
n
3
2
+
⋯
+
11
+
n
1
1
2
)
…
Mathematics
Problem
Let
P
=
∑
n
=
1
11
n
(
1
2
1
+
n
+
2
2
2
+
n
+
3
2
3
+
n
+
⋯
+
11
2
11
+
n
)
P = \sum_{n=1}^{11} n \left( \frac{1^2}{1+n} + \frac{2^2}{2+n} + \frac{3^2}{3+n} + \dots + \frac{11^2}{11+n} \right)
P
=
∑
n
=
1
11
n
(
1
+
n
1
2
+
2
+
n
2
2
+
3
+
n
3
2
+
⋯
+
11
+
n
1
1
2
)
then value of
2
P
\sqrt{2P}
2
P
is
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