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Byju's Answer
Standard XII
Mathematics
Sequence
Find the sum ...
Question
Find the sum to the series
1.
n
+
2
(
n
−
1
)
+
3
(
n
−
2
)
+
.
.
.
+
n
.1
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Solution
Given series:
1.
n
+
2
(
n
−
1
)
+
3
(
n
−
2
)
+
…
.
.
+
n
.1
r
t
h
term
=
r
(
n
−
r
+
1
)
The given series is summation of
r
(
n
−
r
+
1
)
where
r
varies from
1
to
n
.
∴
S
u
m
=
n
∑
r
=
1
[
(
n
+
1
)
r
−
r
2
]
=
(
n
+
1
)
(
n
)
(
n
+
1
)
2
−
n
(
n
+
1
)
(
2
n
+
1
)
6
S
u
m
=
n
(
n
+
1
)
2
[
n
+
1
−
2
n
+
1
3
]
Or,
S
u
m
=
n
(
n
+
1
)
2
[
3
n
+
3
−
2
n
−
1
3
]
∴
S
u
m
=
n
(
n
+
1
)
2
[
n
+
2
3
]
=
n
(
n
+
1
)
(
n
+
2
)
6
Ans.
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Q.
The sum of the series
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