Given, an=6n−5n,n=1,2,3,... We take; 6n=(1+5)n Expand with binomial expansion 6n=nC0+nC15+nC252+nC353+... 6n=1+n⋅5+nC225+nC353+... (6n−5n)=1+25{nC2+nC3⋅5+...} (6n−5n)=1+25⋅k where k= positive integer. Hence, an=6n−5n divided by 25 and leave the remainder =1