Question

последовательность задана формулой bn=3n-4.найти b1,b6,b10,bk,bk+1?

Answer 1

bn=3n-4

b1. n=1, b1=3*1-4=-1;
b6. n=2, b6=6*3-4=18-4=14;
b10. n=10, b10=10*3-4=30-4=26;
bk. n=k, bk=3k-4;
bk+1, n=k+1, bk+1=3(k+1)-4=3k+3-4=3k-1/