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Merge mark calculated

hello friends let's now try to analyze mergesort or try to find the time complexity of Morris or so as we know merge sort is a recursive algorithm so we'll be writing the recurrence for merge sort and we'll be trying to solve that recurrence by using the meta method of repeated substitution so here is the recurrence for merge sort T of n is equal to time taken by merge sort for the input of size n is equal to time taken to sort the left half which is of size n by 2 plus time taken to sort the right half which is also of size n by 2 plus time needed to merge the left half and the right half which if we compute it is exactly or roughly equal to n that means in order to merge this array of size n by 2 with the other area of size n by 2 we'll roughly do n comparisons and so if you remember so merge sort will be given this array a let's say so what we do in merge sort is we compute the left compute the center or we find the center and say this is my left half from L to mid and this is my right half from mid plus 1 to R that means I am doing the division of this array exactly half of that means if originally I have n elements left half will contain n by 2 the right half will contain remaining n by 2 elements so thus we write the recurrence time needed to sort this array of size n is equal to time needed to sort this left half plus time needed to sort this right half plus once I get the left half sorted and right have sorted I need the time to merge these two left off and right half now to get the array entirely sorted so that time equally in if we do this computation that's how I get the recurrence so if I rewrite this recurrence it will be twice T of n by 2 plus n so T of n is equal to twice T of n by 2 plus 1 and so this is equation 1 and this is the recurrence for merge sort so now let's try to solve this by using method of repeated substitution so arguing recurrence means getting rid of this T on the right hand side so now I don't know that this T of n by 2 but I will a value of n by 2 from Equation number 1 silver from the same equation so substitute n by 2 in place of an in equation 1 so equation 1 becomes T of N by 2 we are substituting n by by everywhere in place of him n by 2 is equal to twice T of n by 2 by 2 it's n by 4 plus n by 2 so that's equation...