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Joint Entrance Examination
Mathematics
Sequences and Series
Question 378
Question 378 of 519
Medium
MCQ
The sum
3
1
2
+
5
1
2
+
2
2
+
7
1
2
+
2
2
+
3
2
+
⋯
\dfrac{3}{1^2}+\dfrac{5}{1^2+2^2}+\dfrac{7}{1^2+2^2+3^2}+\cdots
1
2
3
+
1
2
+
2
2
5
+
1
2
+
2
2
+
3
2
7
+
⋯
up to 11 terms is:
A
7
2
\dfrac72
2
7
B
11
4
\dfrac{11}{4}
4
11
C
11
2
\dfrac{11}{2}
2
11
D
60
11
\dfrac{60}{11}
11
60
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Solutions · 1
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Official solution
The
n
n
n
-th term is
T
n
=
2
n
+
1
n
(
n
+
1
)
(
2
n
+
1
)
/
6
=
6
n
(
n
+
1
)
=
6
(
1
n
−
1
n
+
1
)
T_n=\dfrac{2n+1}{n(n+1)(2n+1)/6}=\dfrac{6}{n(n+1)}=6\left(\dfrac1n-\dfrac{1}{n+1}\right)
T
n
=
n
(
n
+
1
)
(
2
n
+
1
)
/6
2
n
+
1
=
n
(
n
+
1
)
6
=
6
(
n
1
−
n
+
1
1
)
. \[S_n=6-\frac{6}{n+1}=\frac{6n}{n+1}.\] \[S_{11}=\frac{6\times11}{12}=\frac{66}{12}=\frac{11}{2}.\]
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