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recurrence relation

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Use substitution method to solve such type of ques

T(n)=T(n-1)+n

T(n-1)=T(n-2)+n-1

Substituting value of T(n-1) we get

T(n)=T(n-2)+n-1+n

Now substituting value of T(n-2) and so on we get

T(n)=T(n-k)+(n-k+1)+.........n-k+n

Put n=k

T(n)=T(0)+1+2..........+n

T(n)=2+n(n+1)/2

 So T(n)=O(n^2)
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T(n)= T(n-1) +n

T(n-1)= T(n-2) +(n-1) [putting n=n-1]

T(n-2)= T(n-3) +(n-2) [again putting n=n-1]

............

T(1)= T(0) +1

----------------------------------------------------

T(n)= T(0) +n+(n-1)+............+1 [Adding all the terms]

T(n)=2+n(n+1)/2

        =2+(n2+n)/2

 T(n)=1/2(n2+n+4)

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