Solve
Guides
Join / Login
Use app
Login
0
You visited us
0
times! Enjoying our articles?
Unlock Full Access!
Standard XI
Mathematics
NCERT
Question
Find an approximation of
(
0.99
)
5
using the first three of its expansion.
0.95
0.97
0.98
89
A
0.98
B
0.95
C
89
D
0.97
Open in App
Solution
Verified by Toppr
The correct option is
A
0.95
(
0.99
)
5
=
(
1
−
0.01
)
5
We know that
(
a
+
b
)
n
=
n
C
0
a
n
+
n
C
1
a
n
−
1
b
1
+
n
C
2
a
n
−
2
b
2
+
−
−
−
−
−
+
n
C
n
−
1
a
1
b
n
−
1
+
n
C
n
b
n
Hence
(
a
+
b
)
5
=
5
C
0
a
5
+
5
C
1
a
4
b
1
+
5
C
2
a
3
b
2
+
5
C
3
a
2
b
2
+
5
C
4
a
b
4
+
5
C
5
b
5
=
a
5
+
5
!
1
!
(
5
−
1
)
!
a
4
b
1
+
5
!
2
!
(
5
−
2
)
!
a
3
b
2
+
5
!
3
!
(
5
−
3
)
!
a
2
b
3
+
5
!
4
!
(
5
−
4
)
!
a
b
4
+
b
5
Using first three terms,
(
0.99
)
5
=
1
−
0.05
+
0.001
=
1.001
−
0.050
=
0.9510
So, approximate value of
(
0.99
)
5
=
0.9510
Was this answer helpful?
3
Similar Questions
Q1
Find an approximation of
(
0.99
)
5
using the first three of its expansion.
View Solution
Q2
Find an approximation of (0.99)
5
using the first three terms of its expansion.