solve The value of 3n-2/r (1+2+3+...+ n), where n is a positive integer, is
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Answer:
know that
(1+x)
3n
=1+
3n
C
1
.x+
3n
C
2
.x
2
+...+
3n
C
3n
.x
3n
...(1)
(1−x)
3n
=1−
3n
C
1
.x+
3n
C
2
.x
2
−...+(−1)
3n
C
3n
.x
3n
...(2)
Subtracting (2) from (1), we get
(1+x)
3n
−(1−x)
3n
=2[
3n
C
1
.x+
3n
C
3
.x
3
+
3n
C
5
.x
5
+...]
=2[
3n
C
1
+
3n
C
3
.x+
3n
C
5
.x
4
+...]
Putting x=i
3
, we get
(1+i
3
)
3n
−(1−i
3
)
3n
=2i
3
[
3n
C
1
−3.
3n
C
3
+3
3
.
3n
C
5
+....]
∴
3n
C
1
−3.
3n
C
3
+3
3
.
3n
C
5
+....=
2i
3
1
[(1+i
3
)
3n
−(1−i
3
)
3n
]
=
2i
3
1
.2
3n
⎣
⎢
⎡
(
2
1
+
2
i
3
)
3n
−(
2
1
−
2
i
3
)
3n
⎦
⎥
⎤
=
i
3
2
3n−1
[(cosnπ+isinnπ)−(cosnπ−isinnπ)]
=
i
3
2
3n−1
2isinnπ=0 as n is an integer
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