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Question

Find the sum to n terms of series 1.2.3+2.3.4+3.4.5+4.5.6+........ .

A
n×(n+1)×(n+2)×(n+3)4
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B
(n+1)×(n+2)×(n+3)4
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C
n×(n+1)×(n+2)×(n+3)12
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D
n×(n+1)×(n+2)×(n+3)8
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Solution

The correct option is A n×(n+1)×(n+2)×(n+3)4
an=n(n+1)(n+2)
Sn=an

=n(n+1)(n+2)

=n(n2+n+2n+2)

=(n3+3n2+2n)

=n3+3n2+2n

=n3+3n2+2n

=n2(n+1)24+3n(n+1)(2n+1)6+2n(n+1)2

=n(n+1)2[n(n+1)2+2n+1+2]

=n(n+1)2[n2+n2+2n+3]

=n(n+1)2[n2+n+4n+62]

=n(n+1)(n2+5n+6)2×2

=n(n+1)(n+2)(n+3)4

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