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Byju's Answer
Standard XII
Mathematics
Sum of Binomial Coefficients with Alternate Signs
If 2nC2+2nC4+...
Question
If
2
n
C
2
+
2
n
C
4
+
2
n
C
6
+
⋯
+
2
n
C
2
n
=
511
, then absolue value of
n
C
0
−
n
C
1
3
1
+
n
C
2
3
2
+
⋯
+
(
−
1
)
n
n
c
n
3
n
is
Open in App
Solution
We know that
2
n
C
0
+
2
n
C
2
+
2
n
C
4
+
2
n
C
6
+
⋯
+
2
n
C
2
n
=
2
2
n
−
1
⇒
512
=
2
2
n
−
1
⇒
n
=
5
Let
S
=
n
C
0
−
n
C
1
3
1
+
n
C
2
3
2
+
⋯
+
(
−
1
)
n
n
C
n
3
n
=
n
∑
r
=
0
(
−
1
)
r
⋅
n
C
r
⋅
3
r
=
n
∑
r
=
0
n
C
r
(
−
3
)
r
=
(
1
−
3
)
n
=
(
−
2
)
n
Since,
n
=
5
∴
S
=
(
−
2
)
5
=
−
32
⇒
|
S
|
=
32
Suggest Corrections
0
Similar questions
Q.
If
2
n
C
2
+
2
n
C
4
+
2
n
C
6
+
⋯
+
2
n
C
2
n
=
511
, then absolue value of
n
C
0
−
n
C
1
3
1
+
n
C
2
3
2
+
⋯
+
(
−
1
)
n
n
c
n
3
n
is
Q.
If
n
C
0
2
n
+
2.
n
C
1
2
n
+
3.
n
C
2
2
n
+
…
.
(
n
+
1
)
n
C
n
2
n
=
16
, then the value of '
n
' is:
Q.
I
f
(
1
+
x
)
n
=
C
0
+
c
1
x
+
.
.
.
+
C
n
x
n
+
,
t
h
e
n
C
1
C
0
+
2
C
2
C
1
+
3
C
3
C
2
+
.
.
.
.
.
+
n
C
n
C
n
−
1
i
s
Q.
If
C
0
,
C
1
,
C
2
,
.
.
.
.
.
.
.
.
.
.
.
C
n
are the Binomial coefficients in the expansion
(
1
+
x
)
n
.
‘n’ being even, then
C
0
+
(
C
0
+
C
1
)
+
(
C
0
+
C
1
+
C
2
)
+
.
.
.
.
.
.
.
.
.
(
C
0
+
C
1
+
C
2
+
.
.
.
.
.
+
C
n
−
1
)
=is equal to
Q.
Prove that
C
0
2
+
C
1
3
+
C
2
4
.
.
.
.
.
.
+
C
n
n
+
2
=
1
+
n
⋅
2
n
+
1
(
n
+
1
)
(
n
+
2
)
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