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JEE Main Math Class 11 Sequences and Series Part 2 Questions
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© examsnet.com
Question : 3
Total: 100
If
1
2
⋅
3
10
+
1
2
2
⋅
3
9
+
.
.
.
+
1
2
10
⋅
3
=
K
2
10
⋅
3
10
, then the remainder when
K
is divided by 6 is :
[25-Jun-2022-Shift-1]
1
2
3
5
Validate
Solution:
👈: Video Solution
1
2
⋅
3
10
+
1
2
2
⋅
3
9
+
.
.
.
+
1
2
10
⋅
3
=
K
2
10
⋅
3
10
⇒
1
2
⋅
3
10
[
(
3
2
)
10
−
1
3
2
−
1
]
=
K
2
10
⋅
3
10
=
3
10
−
2
10
2
10
⋅
3
10
=
K
2
10
⋅
3
10
⇒
K
=
3
10
−
2
10
Now
K
=
(
1
+
2
)
10
−
2
10
=
10
C
0
+
10
C
1
2
+
10
C
2
2
3
+
.
.
.
+
10
C
10
2
10
−
2
10
=
10
C
0
+
10
C
1
2
+
6
λ
+
10
C
9
⋅
2
9
=
1
+
20
+
5120
+
6
λ
=
5136
+
6
λ
+
5
=
6
µ
+
5
λ
,
µ
∈
N
∴
remainder
=
5
© examsnet.com
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