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Byju's Answer
Standard XII
Mathematics
General Term of Binomial Expansion
C 0 + 3 2 . C...
Question
C
0
+
3
2
.
C
1
+
9
3
.
C
2
+
27
4
.
C
3
+
.
.
.
.
.
.
.
.
.
.
.
+
3
n
n
+
1
.
C
n
=
4
n
+
1
−
1
3
(
n
+
1
)
.Is it true ?If true enter 1 else 0.
Open in App
Solution
Solve: we have Given.
c
0
+
3
2
⋅
c
1
+
9
3
c
2
+
27
4
c
3
+
⋯
+
3
n
n
+
1
c
n
=
4
n
+
1
−
1
3
(
n
+
1
)
By binomial expansion
(
1
+
x
)
n
=
n
c
0
+
n
c
1
x
+
n
c
2
x
2
+
…
.
.
n
c
n
x
n
on integrating both sides with respect to
x
we get
∫
x
0
(
1
+
x
)
n
d
x
=
∫
(
n
c
0
+
n
c
1
x
+
n
c
2
x
2
+
…
n
c
n
x
n
)
d
x
⇒
(
1
+
x
)
n
+
1
−
1
n
+
1
=
n
c
0
x
+
n
c
1
x
2
2
+
n
c
2
x
3
3
+
…
n
c
n
x
n
+
1
n
+
1
on putting
x
=
3
we get
⇒
4
n
+
1
−
1
n
+
1
=
3
n
c
0
+
9
2
n
c
1
+
27
3
n
c
2
+
…
.
.
3
n
+
1
n
+
1
n
c
n
⇒
4
n
+
1
−
1
n
+
1
=
3
[
n
c
0
+
3
2
n
c
1
+
9
3
n
c
2
+
…
…
3
n
+
1
n
+
1
n
c
n
]
⇒
n
c
0
+
3
2
n
c
1
+
9
3
n
c
2
+
…
3
n
n
+
1
=
4
n
+
1
−
1
3
(
n
+
1
)
Hence It is correct
so, correct answer is 1 .
Suggest Corrections
0
Similar questions
Q.
C
0
+
3
2
.
C
1
+
C
2
+
27
4
.
C
3
+
+
3
n
n
+
1
.
C
n
=
4
n
+
1
+
1
3
(
n
+
1
)
Q.
Prove that
C
0
+
C
1
2
+
C
2
3
+
C
3
4
+
.
.
.
.
+
C
n
n
+
1
=
2
n
+
1
−
1
(
n
+
1
)
Q.
The value of
2
(
n
C
0
)
+
3
2
(
n
C
1
)
+
4
3
(
n
C
2
)
+
5
4
(
n
C
3
)
.
.
.
is:
Q.
C
1
C
0
+
2.
C
2
C
1
+
3.
C
3
C
2
+
.
.
.
.
.
+
n
.
C
n
C
n
−
1
=
Q.
The sum of the series
2
C
0
+
C
1
2
⋅
2
2
+
C
2
3
⋅
2
3
+
C
3
4
⋅
2
4
+
⋯
+
C
n
n
+
1
⋅
2
n
+
1
is equal to , where
(
C
r
=
n
C
r
)
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