Easy: Solve T(n)=T(n-1)+n by Iteration Method - algorithm

Question

Can someone please help me with this ?
Use iteration method to solve it. T(n) = T(n-1) +n
Explanation of steps would be greatly appreciated.

Answer 1

T(n) = T(n-1) + n
T(n-1) = T(n-2) + n-1
T(n-2) = T(n-3) + n-2
and so on you can substitute the value of T(n-1) and T(n-2) in T(n) to get a general idea of the pattern.
T(n) = T(n-2) + n-1 + n
T(n) = T(n-3) + n-2 + n-1 + n
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T(n) = T(n-k) + kn - k(k-1)/2 ...(1)
For base case:
n - k = 1 so we can get T(1)
=> k = n - 1
substitute in (1)
T(n) = T(1) + (n-1)n - (n-1)(n-2)/2
Which you can see is of Order n2 => O(n2).

Answer 2

Expand it!
T(n) = T(n-1) + n = T(n-2) + (n-1) + n = T(n-3) + (n-2) + (n-1) + n
and so on, until
T(n) = 1 + 2 + ... + n = n(n+1)/2 [= O(n^2)]
provided that T(1) = 1

Answer 3

In pseudo code using iteration:
function T(n) {
int result = 0;
for (i in 1 ... n) {
result = result + i;
}
return result;
}

Answer 4

Another solution:
T(n) = T(n-1) + n
= T(n-2) + n-1 + n
= T(n-3) + n-2 + n-1 + n
// we can now generalize to k
= T(n-k) + n-k+1 + n-k+2 + ... + n-1 + n
// since n-k = 1 so T(1) = 1
= 1 + 2 + ... + n //Here
= n(n-1)/2
= n^2/2 - n/2
// we take the dominating term which is n^2*1/2 therefor 1/2 = big O
= big O(n^2)

Answer 5

Easy Method:
T (n) = T (n - 1) + (n )-----------(1)
//now submit T(n-1)=t(n)
T(n-1)=T((n-1)-1)+((n-1))
T(n-1)=T(n-2)+n-1---------------(2)
now submit (2) in (1) you will get
i.e T(n)=[T(n-2)+n-1]+(n)
T(n)=T(n-2)+2n-1 //simplified--------------(3)
now, T(n-2)=t(n)
T(n-2)=T((n-2)-2)+[2(n-2)-1]
T(n-2)=T(n-4)+2n-5---------------(4)
now submit (4) in (2) you will get
i.e T(n)=[T(n-4)+2n-5]+(2n-1)
T(n)=T(n-4)+4n-6 //simplified
............
T(n)=T(n-k)+kn-6
**Based on General form T(n)=T(n-k)+k, **
now, assume n-k=1 we know T(1)=1
k=n-1
T(n)=T(n-(n-1))+(n-1)n-6
T(n)=T(1)+n^2-n-10
According to the complexity 6 is constant
So , Finally O(n^2)