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Byju's Answer
Standard IX
Mathematics
Coefficients of a Polynomial
Let n b...
Question
Let
n
be a positive integer and define
f
(
n
)
=
1
!
+
2
!
+
3
!
+
⋯
+
n
!
.Find Polynomials
P
(
x
)
and
Q
(
x
)
such that
f
(
n
+
2
)
=
Q
(
n
)
f
(
n
)
+
P
(
n
)
f
(
n
+
1
)
forall
n
≥
1
Find P(2).
Open in App
Solution
f
(
n
)
=
1
!
+
2
!
+
3
!
+
⋯
+
n
!
f
(
n
+
2
)
=
Q
(
n
)
f
(
n
)
+
P
(
n
)
f
(
n
+
1
)
1
!
+
2
!
+
⋯
+
(
n
+
2
)
!
=
Q
(
n
)
[
1
!
+
2
!
+
⋯
+
n
!
]
+
P
(
n
)
[
1
!
+
2
!
+
⋯
+
(
n
+
1
)
!
]
Comparing coefficients upto
(
n
−
1
)
!
Q
(
n
)
+
P
(
n
)
=
1
⟹
Q
(
n
)
=
1
−
P
(
n
)
n
!
+
(
n
+
1
)
!
+
(
n
+
2
)
!
=
(
n
!
)
Q
(
n
)
+
(
n
!
+
(
n
+
1
)
!
)
P
(
n
)
1
+
n
+
1
+
n
2
+
3
n
+
2
=
1
−
P
(
n
)
+
(
n
+
2
)
P
(
n
)
P
(
n
)
=
n
+
3
⟹
Q
(
n
)
=
−
(
n
+
2
)
P
(
2
)
=
5
Suggest Corrections
0
Similar questions
Q.
Let f(n) =
1
+
1
2
+
1
3
+
1
4
+
.
.
.
+
1
n
such that P(n)f(n + 2) = P(n)f(n) + q(n) then match the following List I with List II
Q.
Let
N
be set of positive integers. For all
n
∈
N
,
let
f
n
=
(
n
+
1
)
1
/
3
−
n
1
/
3
and
A
=
{
n
∈
N
;
f
n
+
1
<
1
3
(
n
+
1
)
2
/
3
<
f
n
}
then
Q.
If
f
(
x
)
is a polynomial of degree
n
such that
f
(
0
)
=
0
,
f
(
1
)
=
1
2
,
.
.
.
.
,
f
(
n
)
=
n
n
+
1
, then the value of
f
(
n
+
1
)
is
Q.
Let p(x) be a polynomial such that p(x) - p'(x)=
x
n
, where n is a positive integer. Then p(0) equals.
Q.
Given
f
(
n
)
being a real valued function whose domain is the set of positive integers and that
f
(
n
)
satisfies the following two properties.
f
(
1
)
=
23
;
f
(
n
+
1
)
=
8
+
3
×
f
(
n
)
for
n
≥
1.
f
(
n
)
takes the form
(
a
.
b
n
)
−
c
for n=1,2,3,4....and a,b,c are constants. find the sum of
a
+
b
+
c
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