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JEE Main Math Class 11 Binomial Theorem Part 2 Questions
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© examsnet.com
Question : 2
Total: 115
Let
a
=
1
+
2
C
2
3
!
+
3
C
2
4
!
+
4
C
2
5
!
+
.
.
.
,
b
=
1
+
1
C
0
+
1
C
1
1
!
+
2
C
0
+
2
C
1
+
2
C
2
2
!
+
3
C
0
+
3
C
1
+
3
C
2
+
3
C
3
3
!
+
.
.
.
Then
2
b
a
2
is equal to _______
[4 Apr 2024 Shift 1]
Your Answer:
Validate
Solution:
👈: Video Solution
f
(
x
)
=
1
+
(
1
+
x
)
1
!
+
(
1
+
x
)
2
2
!
+
(
1
+
x
)
3
3
!
+
.
.
.
.
e
(
1
+
x
)
1
+
x
=
1
1
+
x
+
1
+
(
1
+
x
)
2
!
+
(
1
+
x
)
2
3
!
+
(
1
+
x
)
2
4
!
coef
x
2
in RHS :
1
+
2
C
2
3
+
3
C
2
4
+
.
.
.
=
a
coeff.
x
2
in L.H.S.
e
(
1
+
x
+
x
2
2
!
)
.
.
.
(
1
−
x
+
x
2
2
!
.
.
.
.
.
)
is
e
−
e
+
e
2
!
=
a
b
=
1
+
2
1
!
+
2
2
2
!
+
2
3
3
!
+
.
.
.
.
.
.
=
e
2
2
b
a
2
=
8
© examsnet.com
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