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NCERT Class XI Mathematics - Sequences and Series - Solutions

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Question : 104 of 106
Marks: +1, -0
Show that
1×22+2×32+...+n×(n+1)2
12×2+22×3+...+n2×(n+1)
=
3n+5
3n+1
Solution:  
We have
1×22+2×32+...+n×(n+1)2
12×2+22×3+...+n2×(n+1)

∴ nth term is an =
n(n+1)2
n2(n+1)
 =
n3+2n2+n
n2+n3
Hence, the sum to n terms is Sn =
n
Σ
k=1
ak =
n
Σ
k=1
(
k3+2k2+k
k3+k2
)
=
n
Σ
k=1
k3+2
n
Σ
k=1
k2+
n
Σ
k=1
k
n
Σ
k=1
k3+
n
Σ
k=1
k2
 =
[
n(n+1)
2
]2+
2n(n+1)(2n+1)
6
+
n(n+1)
2
[
n(n+1)
2
]2+
n(n+1)(2n+1)
6

=
n(n+1)
2
[
n(n+1)
2
+
2(2n+1)
3
+1]
n(n+1)
2
[
n(n+1)
2
+
(2n+1)
3
]
=
3n2+3n+8n+4+6
3n2+3n+4n+2
 =
3n2+11n+10
3n2+7n+2
=
(3n+5)(n+2)
(3n+1)(n+2)
 =
3n+5
3n+1
Hence,
1×22+2×32+...+n×(n+1)2
12×2+22×3+...+n2×(n+1)
=
3n+5
3n+1
 =
3n+5
3n+1
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