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Byju's Answer
Standard XII
Mathematics
Sum of Coefficients of All Terms
If for 1≤ m...
Question
If for
1
≤
m
≤
n
,
f
(
m
,
n
)
=
C
0
−
C
1
+
C
2
−
.
.
.
.
(
−
1
)
m
−
1
C
m
−
1
, find
f
(
9
,
5
)
.
Open in App
Solution
1
≤
m
≤
n
;
f
(
m
,
n
)
=
C
0
−
C
1
+
C
2
−
.
.
.
.
(
−
1
)
m
−
1
C
m
−
1
f
(
m
,
n
)
is equal to the constant term in the expansion of
=
[
C
0
−
C
1
x
+
C
2
x
2
−
.
.
.
.
.
+
(
−
1
)
n
C
n
x
n
]
[
1
+
1
x
+
1
x
2
+
.
.
.
.
.
.
.
+
1
x
m
−
1
]
=
(
1
−
x
)
n
(
1
−
1
x
m
)
(
1
−
1
x
)
=
(
1
−
x
)
n
−
1
(
1
−
x
m
)
f
(
m
,
n
)
=
Coefficient of
x
m
−
1
in
(
1
−
x
)
n
−
1
(
1
−
x
m
)
f
(
m
,
n
)
=
Coefficient of
x
m
−
1
in
(
1
−
x
)
n
−
1
General Term of the expansion
(
1
−
x
)
n
is
T
r
+
1
=
(
−
1
)
r
(
n
C
r
x
r
)
f
(
m
,
n
)
=
(
−
1
)
m
−
1
(
n
−
1
C
m
−
1
)
f
(
9
,
5
)
=
(
−
1
)
5
−
1
(
9
−
1
C
5
−
1
)
=
8
C
4
=
70.
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0
Similar questions
Q.
If
(
1
+
x
)
n
=
C
0
+
C
1
x
+
C
2
x
2
+
⋯
+
C
n
x
n
,
n
∈
N
,
then for
2
≤
m
≤
n
,
C
0
−
C
1
+
C
2
−
⋯
+
[
(
−
1
)
m
−
1
]
C
m
−
1
is equal to
Q.
Assertion :
1
m
!
C
0
+
n
(
m
+
1
)
!
C
1
+
n
(
n
−
1
)
(
m
+
2
)
!
C
2
+
.
.
.
.
.
+
n
(
n
−
1
)
.
.
.2
.1
(
m
+
n
)
!
C
n
=
(
m
+
n
+
1
)
(
m
+
n
+
2
)
.
.
.
(
m
+
2
n
)
(
m
+
n
)
!
Reason: for
r
≥
0
(
m
r
)
C
0
+
(
m
r
−
1
)
C
1
+
.
.
.
.
.
.
(
m
0
)
C
r
=
(
m
+
n
r
)
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.
If
n
C
r
=
C
r
then
C
0
+
(
C
0
+
C
1
)
+
(
C
0
+
C
1
+
C
2
)
+
.
.
.
.
+
(
C
0
+
C
1
+
.
.
.
.
.
+
C
n
)
=
Q.
If
(
1
+
x
2
)
2
(
1
+
x
)
n
=
C
0
+
C
1
x
+
C
2
x
2
+
⋅
⋅
⋅
,
and if
C
0
,
C
1
,
C
2
are in A.P., find
n
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