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Byju's Answer
Standard XII
Mathematics
Director Circle: Ellipse
n0+2n1+22n2+....
Question
(
n
0
)
+
2
(
n
1
)
+
2
2
(
n
2
)
+
.
.
.
.
.
+
2
n
(
n
n
)
is equal to.
A
2
n
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B
0
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C
3
n
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D
None of these
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Solution
The correct option is
C
3
n
We know that the expansion of
(
x
+
1
)
n
is
=
(
n
0
)
+
x
(
n
1
)
+
x
2
(
n
2
)
+
⋯
+
x
n
(
n
n
)
If we put
x
=
2
in the above equation then,
(
2
+
1
)
n
=
(
n
0
)
+
2
(
n
1
)
+
⋯
+
2
n
(
n
n
)
=
3
n
Suggest Corrections
0
Similar questions
Q.
(
n
0
)
+
2
(
n
1
)
+
2
2
(
n
2
)
+
.
.
.
.
+
2
n
(
n
n
)
is equal to
Q.
lim
n
→
∞
5
n
+
1
+
3
n
−
2
2
n
5
n
+
2
n
+
3
2
n
+
3
,
n
∈
N
is equal to
Q.
The value of
C
0
+
3
C
1
+
5
C
2
+
7
C
3
+
.
.
.
.
+
(
2
n
+
1
)
C
n
is equal to :
Q.
If
(
1
+
x
)
n
=
n
∑
r
=
0
n
C
r
x
n
,
then
C
0
1
⋅
2
2
2
+
C
1
2
⋅
3
2
3
+
C
2
3
⋅
4
2
4
+
⋯
+
C
n
(
n
+
1
)
(
n
+
2
)
2
n
+
2
is equal to
Q.
If
(
1
+
x
)
n
=
C
0
+
C
1
x
+
C
2
x
2
+
…
+
C
n
x
n
,
then
the value of
∑
∑
0
≤
r
<
s
≤
n
(
r
+
s
)
C
r
C
s
is
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Standard XII Mathematics
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