2.1.2 Recurrence Relation (T(n)= T(n-1) + n) #2

Recurrence Relation for Decreasing Function
Example : T(n)= T(n-1) +n
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Excellent teacher. As he said, adding the n gives 2n-1, 3n-2, etc, making the pattern a bit less obvious and the expression a bit more difficult to solve. But sometimes we have to live dangerously, so here we go: T(n) = T(n-1) + n T(n) = T(n-2) + 2n - 1 T(n) = T(n-3) + 3n - 3 T(n) = T(n-4) + 4n - 6 ... T(n) = T(n-k) + kn - [(k-1) + (k-2) + ... + 3 + 2 + 1] Assuming n-k = 0; k = n T(n) = T(0) + n² - [(n-1) + (n-2) + ... + 3 + 2 + 1] Isolating the expression in brackets [(n-1) + (n-2) + ... + 2 + 1] We know that n + (n-1) + (n-2) + ... + 2 + 1 = n(n+1)/2 Taking off n from both sides we get (n-1) + (n-2) + ... + 2 + 1 = (n(n+1)/2) - n which resolves to (n(n+1)/2) - 2n/2 => (n(n+1) - 2n)/2 => (n(n+1-2))/2 => n(n-1)/2 Plugging it back in the equation in its new form T(n) = T(0) + n² - [n(n-1)/2] Removing the brackets and swapping the sign T(n) = T(0) + n² + n(1-n)/2 T(n) = 1 + (2n² + n(1-n))/2 T(n) = 1 + n(2n+1-n)/2 T(n) = 1 + n(n+1)/2

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its T(n) = T(n-3)+3n-3 I think

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I would like to ask, where did 2n^2 came from after removing and swapping the sign? did you just plugged in a constant c to get n(n+1)/2?

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Should mention that 1 + 2 + 3 + .... n-1 + n is a summation that can be shorthanded to equal n(n+1)/2

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Sir, ar 5:51 you took T(0) will take 0 time. But you also said that 0 time means constant time so we take 1 instead of 0. So, for summing up, if taken 1 instead of 0 sum should be = n(n+1)/2 +1 . I know it wont make a difference in the final answer but just wanted to confirm if I am theoritically correct or not ? Please do reply and Thank you in advance !

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A probable rectification - Since T(0) is never written as 0 as instructed by you in the previous video , T(0) in the execution depiction should be equal to 1not 0. So it should be 1+2+3+......+ n-1+n = n(n-1)/2 .

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