If (4+15)n=I+f, where n is an odd natural number, I is an integer and 0 < f < 1, then

I+f=(4+15)nLet f'=(4+15)n. Then 0 < f' < 1I+f=nC04n+nC14n−115+nC24n−215+nC34n−3(15)3+⋯f'=nC04n−nC14n−115+nC24n−2⋅15−nC34n−3(15)3+⋯∴ I+f+f'=2 nC04n+nC24n−2×15+⋯=even integer∵ 0