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Byju's Answer
Standard XI
Mathematics
Summation by Sigma Method
The value of ...
Question
The value of
3
×
1
3
1
2
+
5
×
(
1
3
+
2
3
)
1
2
+
2
2
+
7
×
(
1
3
+
2
3
+
3
3
)
1
2
+
2
2
+
3
2
+
⋯
upto
10
terms
, is
A
600
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B
620
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C
660
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D
680
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Solution
The correct option is
C
660
Let
S
=
3
×
1
3
1
2
+
5
×
(
1
3
+
2
3
)
1
2
+
2
2
+
7
×
(
1
3
+
2
3
+
3
3
)
1
2
+
2
2
+
3
2
+
⋯
upto
10
terms
So, the general term is
T
n
=
(
2
n
+
1
)
(
1
3
+
2
3
+
3
3
+
⋯
+
n
3
)
(
1
2
+
2
2
+
3
2
+
⋯
+
n
2
)
=
(
2
n
+
1
)
(
n
(
n
+
1
)
2
)
2
n
(
n
+
1
)
(
2
n
+
1
)
6
=
3
n
(
n
+
1
)
2
Now,
S
=
10
∑
n
=
1
T
n
=
3
2
[
10
∑
n
=
1
n
2
+
10
∑
n
=
1
n
]
=
3
2
[
n
(
n
+
1
)
(
2
n
+
1
)
6
+
n
(
n
+
1
)
2
]
=
3
2
[
10
⋅
(
11
)
⋅
(
21
)
6
+
10
⋅
(
11
)
2
]
∴
S
=
660
Suggest Corrections
0
Similar questions
Q.
The value of
3
×
1
3
1
2
+
5
×
(
1
3
+
2
3
)
1
2
+
2
2
+
7
×
(
1
3
+
2
3
+
3
3
)
1
2
+
2
2
+
3
2
+
⋯
upto
10
terms
, is
Q.
The sum
3
×
1
3
1
2
+
5
×
(
1
3
+
2
3
)
1
2
+
2
2
+
7
×
(
1
3
+
2
3
+
3
3
)
1
2
+
2
2
+
3
2
+
⋯
upto
10
th
term, is :
Q.
1
3
+
1
2
+
1
+
2
3
+
2
2
+
2
+
3
3
+
3
2
+
3
+
.
.
.
3
n
terms
=
Q.
1
3
+
2
3
+
3
3
+
4
3
+
.
.
.
.
.
.
.
.12
3
1
2
+
2
2
+
3
2
+
4
2
+
.
.
.
.
.
.12
2
=
Q.
Assertion :If
t
r
=
1
2
+
2
2
+
3
2
+
.
.
.
+
r
2
1
3
+
2
3
+
.
.
.
+
r
3
and
S
n
=
n
∑
r
=
1
(
−
1
)
r
.
t
r
,
then
lim
n
→
∞
S
n
=
2
3
Reason:
1
2
+
2
2
+
3
2
+
.
.
.
+
r
2
=
r
(
r
+
1
)
(
2
r
+
1
)
6
and
1
3
+
2
3
+
3
3
+
.
.
.
+
r
3
=
(
r
(
r
+
1
)
2
)
2
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