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Standard XI
Mathematics
NCERT
Question
Using Binomial theorem, evaluate
(
99
)
5
Open in App
Solution
Verified by Toppr
(
99
)
5
=
(
100
−
1
)
5
=
5
C
0
(
100
)
5
−
5
C
1
(
100
)
4
(
1
)
+
5
C
2
(
100
)
3
(
1
)
2
−
5
C
3
(
100
)
2
(
1
)
3
+
5
C
4
(
100
)
(
1
)
4
−
5
C
5
(
1
)
5
=
(
100
)
5
−
5
(
100
)
4
+
10
(
100
)
3
−
10
(
100
)
2
+
5
(
100
)
−
1
=
10000000000
−
500000000
+
10000000
−
100000
+
500
−
1
=
10010000500
−
500100001
=
9509900499
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