binomial approxmation of 3√(999) will be
Answers
Answer:
Answer:
Rewrite
999
3
as
(
10
3
−
1
)
3
and apply the binomial theorem to find that
999
3
=
997002999
Explanation:
The binomial theorem states that
(
a
+
b
)
n
=
n
∑
k
=
0
(
n
k
)
a
n
−
k
b
k
where
(
n
k
)
=
n
!
k
!
(
n
−
k
)
!
For this problem, we will only need to calculate for
n
=
3
, and we will find that
(
3
0
)
=
(
3
3
)
=
1
and
(
3
1
)
=
(
3
2
)
=
3
(Try verifying this)
Noting that it is much easier to calculate powers of
10
and
1
compared to
999
, we can rewrite
999
as
1000
−
1
=
10
3
−
1
and apply the binomial theorem:
999
3
=
(
10
3
−
1
)
3
=
(
3
0
)
(
10
3
)
3
+
(
3
1
)
(
10
3
)
2
(
−
1
)
+
(
3
2
)
(
10
3
)
(
−
1
)
2
+
(
3
3
)
(
−
1
)
3
=
10
9
−
3
⋅
10
6
+
3
⋅
10
3
−
1
=
1000000000
−
3000000
+
3000
−
1
=
997002999Answer:
Rewrite
999
3
as
(
10
3
−
1
)
3
and apply the binomial theorem to find that
999
3
=
997002999
Explanation:
The binomial theorem states that
(
a
+
b
)
n
=
n
∑
k
=
0
(
n
k
)
a
n
−
k
b
k
where
(
n
k
)
=
n
!
k
!
(
n
−
k
)
!
For this problem, we will only need to calculate for
n
=
3
, and we will find that
(
3
0
)
=
(
3
3
)
=
1
and
(
3
1
)
=
(
3
2
)
=
3
(Try verifying this)
Noting that it is much easier to calculate powers of
10
and
1
compared to
999
, we can rewrite
999
as
1000
−
1
=
10
3
−
1
and apply the binomial theorem:
999
3
=
(
10
3
−
1
)
3
=
(
3
0
)
(
10
3
)
3
+
(
3
1
)
(
10
3
)
2
(
−
1
)
+
(
3
2
)
(
10
3
)
(
−
1
)
2
+
(
3
3
)
(
−
1
)
3
=
10
9
−
3
⋅
10
6
+
3
⋅
10
3
−
1
=
1000000000
−
3000000
+
3000
−
1
=
997002999
Step-by-step explanation: