Hi, I wanted to confirm, is this the correct way to solve the following recurren

Question

Hi,

I wanted to confirm, is this the correct way to solve the following recurrence relations? 7T(n/7)+n^2

i = 0 to i = log7n
SIGMA [(7^i)(n/7^i) + i*n^2] = n + (n^2)*(log7n) = (n^2)*(log7n) 49T(n/7)+n^2

i = 0 to i = log7n
SIGMA [(49^i)(n/7^i) + i*n^2] = n^2 + (n^2)*(log7n) =  (n^2)log(7n) 49T(n/7)+n

i = 0 to i = log7n
SIGMA [(49^i)(n/7^i) + i*n] = n^2 + n(log7n) = n^2 and 48T(n/7) + n i = 0 to i = log7n assume logn = log7n SIGMA [(48^i)(n/7^i) + i*n] = n*(n^log(48/7))+nlogn = n*(n^log(48/7)) I wanted to confirm, is this the correct way to solve the following recurrence relations? SIGMA [(7^i)(n/7^i) + i*n^2] = n + (n^2)*(log7n) = (n^2)*(log7n) 49T(n/7)+n^2

i = 0 to i = log7n
SIGMA [(49^i)(n/7^i) + i*n^2] = n^2 + (n^2)*(log7n) =  (n^2)log(7n) 49T(n/7)+n

i = 0 to i = log7n
SIGMA [(49^i)(n/7^i) + i*n] = n^2 + n(log7n) = n^2 and 48T(n/7) + n i = 0 to i = log7n assume logn = log7n SIGMA [(48^i)(n/7^i) + i*n] = n*(n^log(48/7))+nlogn = n*(n^log(48/7))

Explanation / Answer

Yes it is correct way to solve this question.

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