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Question

Sum to n terms of the series 11.2.3+32.3.4+53.4.5+74.5.6+....is

A
n(n+1)2(n+2)(n+3)
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B
n(3n+5)4(n+1)(n+2)
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C
165(n+1)(n+4)
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D
n(3n+1)4(n+1)(n+2)
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Solution

The correct option is D n(3n+1)4(n+1)(n+2)
11.2.3+32.3.4+53.4.5+74.5.6+....uptonterms
=nr=12r1r(r+1)(r+2)
=nr=12(r+1)(r+2)1r(r+1)(r+2)
=2nr=1(r+2)(r+1)(r+1)(r+2)12(nr=11r(r+1)1(r+1)(r+2))
=2nr=1(1r+11r+2)12nr=1(1r1r+1)+12nr=1(1r+11r+2)
=2(121n+2)12(11n+1)+12(121n+2)
=nn+2n2(n+1)+n4(n+2)
After simplifying, we get
=n(3n+1)4(n+1)(n+2)
Hence, option 'D' is correct.

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