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Byju's Answer
Standard XII
Mathematics
Middle Terms
The coefficie...
Question
The coefficient of
x
k
in the expansion of
E
=
1
+
(
1
+
x
)
+
(
1
+
x
)
2
+
.
.
.
.
+
(
1
+
x
)
n
is
A
n
C
k
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B
n
+
1
C
k
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C
n
+
1
C
k
+
1
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D
none of these
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Solution
The correct option is
C
n
+
1
C
k
+
1
Given,
E
=
1
+
(
1
+
x
)
+
(
1
+
x
)
2
+
.
.
.
(
1
+
x
)
n
=
1
−
(
1
+
x
)
n
+
1
1
−
(
1
+
x
)
=
1
x
[
(
1
+
x
)
n
+
1
−
1
]
Hence,
x
k
=
x
k
+
1
x
Coefficient of
x
k
will be
n
+
1
C
k
+
1
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1
Similar questions
Q.
The expansion of
(
1
+
x
)
n
has 3 consecutive terms with coefficient in the ratio
1
:
2
:
3
and can be written in the form
n
C
k
;
n
C
k
+
1
:
n
C
k
+
2
.
. The sum of all possible values of
n
+
k
is
Q.
Assertion :If
n
is a positive integer and
k
is a positive integer not exceeding
n
, then
n
∑
k
=
1
k
3
.
(
C
k
C
k
−
1
)
2
, where
C
k
=
n
C
k
, is
n
(
n
+
1
)
2
(
n
+
2
)
12
Reason:
C
k
C
k
−
1
=
n
C
k
n
C
k
−
1
=
n
−
k
+
1
k
Q.
The value of the expression
k
−
1
C
k
−
1
+
k
C
k
−
1
+
.
.
.
.
n
+
k
−
2
C
k
−
1
is given by :
Q.
Coefficient of
x
k
,
(
0
≤
k
≤
n
)
in expansion of
P
=
1
+
(
1
+
x
)
+
(
1
+
x
)
2
.
.
.
.
.
.
+
(
1
+
x
)
n
Q.
Let
A
k
=
n
C
k
n
C
k
+
n
C
k
+
1
. If
3
√
∑
n
k
=
0
A
k
=
4
then
n
equals
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