Problem:
By definition r!=r(r−1)⋯1 and (kj​)=k!(j−k)!j!​, where r,j,k are positive integers and k<j. If (1n​),(2n​),(3n​) form an arithmetic progression with n>3, then n equals
Answer Choices:
A. 5
B. 7
C. 9
D. 11
E. 12
Solution:
Write
(1n​)=n,(2n​)=2n(n−1)​ and (3n​)=6n(n−1)(n−2)​.
Hence
2n(n−1)​−n=6n(n−1)(n−2)​−2n(n−1)​
Thus
​n3−9n2+14n=0n(n−2)(n−7)=0​
Since n>3,n=7 is the solution.
(The answer may also be obtained by evaluating the sequence (1n​),(2n​),(3n​) for the values of n listed as choices.)