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Tardigrade
Question
Mathematics
If S n denote the sum of n terms of the series 1 ⋅ 2+2 ⋅ 3+3 ⋅ 4+ ldots ldots ldots ldots and σn-1 that to (n-1) terms of the series (1/1 ⋅ 2 ⋅ 3 ⋅ 4)+(1/2 ⋅ 3 ⋅ 4 ⋅ 5)+(1/3 ⋅ 4 ⋅ 5 ⋅ 6)+ ldots ldots ldots ldots . . .. Find |Sn(18 σn-1-1)|.
Q. If
S
n
denote the sum of
n
terms of the series
1
⋅
2
+
2
⋅
3
+
3
⋅
4
+
…………
and
σ
n
−
1
that to
(
n
−
1
)
terms of the series
1
⋅
2
⋅
3
⋅
4
1
+
2
⋅
3
⋅
4
⋅
5
1
+
3
⋅
4
⋅
5
⋅
6
1
+
…………
...
.
Find
∣
S
n
(
18
σ
n
−
1
−
1
)
∣
.
93
141
Sequences and Series
Report Error
Answer:
2
Solution:
S
n
=
1
⋅
2
+
2
⋅
3
+
3
⋅
4
+
………
upto
n
terms
=
n
=
1
∑
n
n
(
n
+
1
)
=
n
=
1
∑
n
n
2
+
n
=
1
∑
n
n
=
6
n
(
n
+
1
)
(
2
n
+
1
)
+
2
n
(
n
+
1
)
S
n
=
3
n
(
n
+
1
)
(
n
+
2
)
Also,
σ
n
−
1
=
1
⋅
2
⋅
3
⋅
4
1
+
2
⋅
3
⋅
4
⋅
5
1
+
3
⋅
4
⋅
5
⋅
6
1
+
……
..
. upto
(
n
−
1
)
terms
=
n
=
1
∑
n
−
1
n
(
n
+
1
)
(
n
+
2
)
(
n
+
3
)
1
=
3
1
n
=
1
∑
n
−
1
n
(
n
+
1
)
(
n
+
2
)
(
n
+
3
)
(
n
+
3
)
−
n
σ
n
−
1
=
3
1
∑
n
=
1
n
−
1
(
n
(
n
+
1
)
(
n
+
2
)
1
−
(
n
+
1
)
(
n
+
2
)
(
n
+
3
)
1
)
⇒
σ
n
−
1
=
18
1
−
3
n
(
n
+
1
)
(
n
+
2
)
1
⇒
σ
n
−
1
=
18
1
−
9
S
n
1
[using(1)]
⇒
σ
n
−
1
=
18
S
n
S
n
−
2
⇒
18
S
n
σ
n
−
1
−
S
n
+
2
=
0
Hence,
∣
S
n
(
18
σ
n
−
1
−
1
)
∣
=
∣
−
2∣
=
2
For objective: Put
n
=
2