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Joint Entrance Examination
Mathematics
Limits, Continuity and Differentiability
Question 280
Question 280 of 401
Medium
MCQ
If
f
(
x
)
=
x
n
f(x)=x^n
f
(
x
)
=
x
n
, then the value of
f
(
1
)
−
f
′
(
1
)
1
!
+
f
′
′
(
1
)
2
!
−
f
′
′
′
(
1
)
3
!
+
⋯
+
(
−
1
)
n
f
(
n
)
(
1
)
n
!
f(1)-\dfrac{f'(1)}{1!}+\dfrac{f''(1)}{2!}-\dfrac{f'''(1)}{3!}+\cdots+\dfrac{(-1)^nf^{(n)}(1)}{n!}
f
(
1
)
−
1
!
f
′
(
1
)
+
2
!
f
′′
(
1
)
−
3
!
f
′′′
(
1
)
+
⋯
+
n
!
(
−
1
)
n
f
(
n
)
(
1
)
is
A
1
B
2
n
2^n
2
n
C
2
n
−
1
2^{n-1}
2
n
−
1
D
0
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Solutions · 1
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Official solution
f
(
k
)
(
1
)
=
n
!
(
n
−
k
)
!
f^{(k)}(1)=\dfrac{n!}{(n-k)!}
f
(
k
)
(
1
)
=
(
n
−
k
)!
n
!
. The sum equals
∑
k
=
0
n
(
−
1
)
k
(
n
k
)
=
(
1
−
1
)
n
=
0
\displaystyle\sum_{k=0}^n(-1)^k\binom nk=(1-1)^n=0
k
=
0
∑
n
(
−
1
)
k
(
k
n
)
=
(
1
−
1
)
n
=
0
.
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