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JEE Main Math Class 11 Sequences and Series Part 3 Questions
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© examsnet.com
Question : 3
Total: 101
The value of
1
×
2
2
+
2
×
3
2
+
.
.
.
+
100
×
(
101
)
2
1
2
×
2
+
2
2
×
3
+
.
.
.
+
100
2
×
101
is
[4 Apr 2024 Shift 2]
306
305
305
301
32
31
31
30
Validate
Solution:
👈: Video Solution
1
×
2
2
+
2
×
3
2
+
.
.
.
+
100
×
(
101
)
2
1
2
×
2
+
2
2
×
3
+
.
.
.
+
100
2
×
101
=
100
Σ
r
=
1
r
(
r
+
1
)
2
100
Σ
r
=
1
r
2
(
r
+
1
)
=
100
Σ
r
=
1
(
r
3
+
2
r
2
+
r
)
100
Σ
r
=
1
(
r
3
+
r
2
)
=
(
n
(
n
+
1
)
2
2
)
+
2
n
(
n
+
1
)
(
2
n
+
1
)
6
+
n
(
n
+
1
)
2
(
n
(
n
+
1
)
2
)
2
+
n
(
n
+
1
)
(
2
n
+
1
)
6
=
n
(
n
+
1
)
2
[
n
(
n
+
1
)
2
+
2
3
⋅
(
2
n
+
1
)
+
1
]
n
(
n
+
1
)
2
[
n
(
n
+
1
)
2
+
(
2
n
+
1
)
3
]
;
Put
n
=
100
=
100
(
101
)
2
+
2
3
(
201
)
+
1
100
×
101
2
+
201
3
=
5185
5117
=
305
301
© examsnet.com
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