Cell Mitosis : Ignoring the case in the odd position

#1

Sir, in this question when we are at odd position, then for calculation of the next odd+1 position, we need the value of the current odd position which we don’t have, and according to the approach, we are just ignoring that option and simply looking for the (odd+1) / 2 value only. How can we just ignore this case, please explain.

#2

we are taking (odd+1)/2 for calculation the cost at position i which is odd not i+1.

#3

Sir, please explain the case when i is odd ? as , in the video bhaiya says that , when i is odd then recurrence relation depends on (i-1) and (i+1) since i/ 2 is not possible , again , for (i+1) its said , that we need simply (i+1)/2 .How is that possible?

#4

I can not able to understand ur exact problem. Pls elaborate more

#5

Sir please explain the relation for odd case. Then only I’ll be able to clearly ask my question.