Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
the series 1+(1/5)+(1 ⋅ 3/5 ⋅ 10)+(1 ⋅ 3 ⋅ 5/5 ⋅ 10 ⋅ 15)+ ldots is equal to:
Q. the series
1
+
5
1
+
5
⋅
10
1
⋅
3
+
5
⋅
10
⋅
15
1
⋅
3
⋅
5
+
…
is equal to:
1935
251
Bihar CECE
Bihar CECE 2003
Report Error
A
5
1
B
2
1
C
3
D
3
5
Solution:
Let
S
=
1
+
5
1
+
5
⋅
10
1
⋅
3
+
5
⋅
10
⋅
15
1
⋅
3
⋅
5
+
…
.
We know that
(
1
+
x
)
n
=
1
+
1
!
n
x
+
2
!
n
(
n
−
1
)
x
2
+
3
!
n
(
n
−
1
)
(
n
−
2
)
x
3
+
…
⇒
n
x
=
5
1
and
2
!
n
(
n
−
1
)
x
2
=
5
⋅
10
1
⋅
3
⇒
n
=
−
2
1
and
x
=
−
5
2
∴
S
=
(
1
−
5
2
)
−
1/2
=
(
5
3
)
−
1/2
=
3
5