If f : N Z is defined by f(n)= vmatrixn & -1 & -5\ -2n2 & 3(2k+1) & 2k+1\ -3n3 & 3k(2k+1) & 3k(k+2)+1 vmatrix, k N, and n=1k f(n)=98, then k is equal to :
- 3
- 4
- 5
- 6
Answer
(1) 3
Solution
Only the first column of f(n) depends on n; columns 2 and 3 are fixed for a given k.
n=1k f(n)= vmatrix n & -1 & -5\ -2 n2 & 3(2k+1) & 2k+1\ -3 n3 & 3k(2k+1) & 3k(k+2)+1 vmatrix
where n= k(k+1)2, n2= k(k+1)(2k+1)6, n3= k2(k+1)24
Cofactors of the first column:
C11=3(2k+1)\3k(k+2)+1\-3k(2k+1)2=3(2k+1)(k2+5k+1)
C21=-\-(3k2+6k+1)+15k(2k+1)\=-(27k2+9k-1)
C31=-(2k+1)+15(2k+1)=14(2k+1)
f(n)= k(k+1)2C11- k(k+1)(2k+1)3C21- 3k2(k+1)24C31
=k(k+1)(2k+1) [ 3(k2+5k+1)2+ 27k2+9k-13- 21k(k+1)2 ]
=k(k+1)(2k+1) (9k2+45k+9)+(54k2+18k-2)-(63k2+63k)6
= 7k(k+1)(2k+1)6
f(n)=98 k(k+1)(2k+1)6=14
k(k+1)(2k+1)=84=3 4 7
k=3
Answer: (1)