Let the smallest value of k N, for which the coefficient of x3 in (1+x)3+(1+x)4+(1+x)5+ +(1+x)99+(1+kx)100, x 0, is (43n+ 1014 ) (100C3 ) for some n N, be p. Then the value of p+n is :
- 10
- 11
- 12
- 13
Answer
(2) 11
Solution
Coefficient of x3= (3C3+4C3+ +99C3 )+k3 100C3
r=399rC3=100C4 (hockey-stick identity) and 100C4= 974 100C3
Coefficient = ( 974+k3 )100C3
974+k3=43n+ 1014 k3-1=43n
(k-1)(k2+k+1) must be a positive multiple of 43 (k=1 gives n=0 N).
k-1 is a multiple of 43 only for k 44; but k2+k+1=43 for k=6.
Checking k=2,3,4,5: k3-1=7, 26, 63, 124, none divisible by 43.
p=6, n= 63-143= 21543=5
p+n=11
Answer: (2)