Find P + 1 if
Pk = 7 + 13 + 19 + ...+[6(k - 1)+1] + (6k + 1)
(the k+1 and k in front of the P are subscripts)
A. 7 + 13 + 19 + …+[6(k - 1) + 1] + (6k + 1) + [6(k + 1) + 1]
B. 8 + 14 + 20 + …+[7(k - 1) + 1] + (7k + 1)
C. 7 + 13 + 19 + …+(6k + 1)
D. 7 + 13 + 19 + ...+[6(k - 1) + 1] + (6k +1) + (6k + 2)
Thank you
well, just add on a k+1 term.
Looks like (A) to me