Divisibility is just an excuse.

Calculus Level 3

If 1 + ∑ r = 0 18 ( r ( r + 2 ) + 1 ) r ! = k ! 1+ \displaystyle \sum_{r=0}^{18} \big(r(r+2)+1 \big)r! =k! then k k is not divisible by

You can try my other Sequences And Series problems by clicking here : Part II and here : Part I.
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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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1 solution

Ronak Agarwal
Feb 11, 2015

S = 1 + ∑ r = 0 18 { r ( r + 2 ) + 1 } . r ! S=1+ \displaystyle \sum_{r=0}^{18} \{ r(r+2) +1 \}.r!

S = 1 + ∑ r = 0 18 { ( r + 1 ) ( r + 2 ) − ( r + 1 ) } . r ! S=1+ \displaystyle \sum_{r=0}^{18} \{ (r+1)(r+2) - (r+1) \}.r!

S = 1 + ∑ r = 0 18 { ( r + 2 ) ! − ( r + 1 ) ! } S=1+ \displaystyle \sum_{r=0}^{18} \{ (r+2)!-(r+1)! \}

Now it's a clear telescoping series.

S = 1 + ( 3 ! − 2 ! + 4 ! − 3 ! + 5 ! − 4 ! + . . . . . . . . . + 20 ! − 19 ! ) = 1 + 20 ! − 1 = 20 ! S=1+ (3!-2! + 4!-3! + 5!-4! +......... + 20!-19! ) = 1+20! -1 = 20!

⇒ k = 20 \Rightarrow k=20

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