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Byju's Answer
Standard XII
Mathematics
Summation by Sigma Method
Prove that ...
Question
Prove that
7
n
{
1
+
n
7
+
n
(
n
−
1
)
7.14
+
n
(
n
−
1
)
(
n
−
2
)
7.14.21
+
.
.
.
.
}
=
4
n
{
1
+
n
2
+
n
(
n
−
1
)
2.4
+
n
(
n
−
1
)
(
n
−
2
)
2.4.6
+
.
.
.
.
}
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Solution
R
H
S
=
4
n
[
1
−
1
2
]
−
n
=
8
n
where n is positive
&
L
H
S
=
7
n
[
1
+
1
7
]
n
=
7
n
[
8
7
]
n
=
8
n
⇒
L
H
S
=
R
H
S
hence proved
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Similar questions
Q.
If
n
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(1)
n
n
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n
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n
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n
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Q.
Prove that
1
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2
n
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3.6
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n
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n
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.
.
.
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n
{
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3.6
+
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1
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n
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.
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Q.
n
n
−
n
(
n
−
1
)
n
+
n
(
n
−
1
)
2
!
(
n
−
2
)
n
−
.
.
.
∞
=
3
a
×
n
!
.
Find
a
Q.
Prove that:
1
1.2.3
+
1
2.3.4
+
1
3.4.5
+
.
.
.
+
1
n
(
n
+
1
)
(
n
+
2
)
=
n
(
n
+
3
)
4
(
n
+
1
)
(
n
+
2
)
.
Q.
Prove that:
n
!
(
n
−
r
)
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
n
−
(
r
−
1
)
)
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