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Question

Sum to n terms of the series 15!+1!6!+2!7!+3!8!+..... is

A
25!1(n+1)!
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B
14(14!n!(n+4)!)
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C
14(13!3!(n+2)!)
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D
14(14!+n!(n+4)!)
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Solution

The correct option is B 14(14!n!(n+4)!)
We have tr=(r1)!(r+4)! and tr+1=r!(r+5)!

Now, rtr(r+5)tr+1=r!(r+4)!r!(r+4)!=0

rtr(r+1)tr+1=4tr+14n1r=1tr+1=n1r=1[rtr(r+1)tr+1]

4(t2+t3+....+tn)=1t1ntn

4(t1+t2+.....tn)=5t1ntn=5(0!5!)n(n1)!(n+4)!

t1+t2+....tn=14[14!n!(n+4)!]

Hence, option 'B' is correct.

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