Let = r=0n (4r2 + 2r + 1 )nCr and = ( r=0n nCrr + 1 ) + 1n + 1. If 140 < 2 < 281, then the value of n is .
Answer
5
Solution
Use nCr = 2n, r nCr = n 2n-1, r2 nCr = n(n + 1)2n-2
= 4n(n + 1)2n-2 + 2n 2n-1 + 2n = 2n [n(n + 1) + n + 1 ] = 2n(n + 1)2
nCrr + 1 = n+1Cr+1n + 1 r=0n nCrr + 1 = 2n+1 - 1n + 1
= 2n+1 - 1 + 1n + 1 = 2n+1n + 1
2 = 2n+1(n + 1)2 (n + 1)2n+1 = (n + 1)3
140 < (n + 1)3 < 281, and the only cube in this range is 216 = 63 (53 = 125, 73 = 343)
So n + 1 = 6, i.e. n = 5
Answer: 5