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

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

substituting 2 in 1 we get

T(n)=T(n-2)+1+1 ...3

T(n-2)=T(n-3)+1   ....4

Substituting 3 in 4

T(n)=T(n-3)+3

so on

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

Basis step is T(1)=1

n-k=1

k=n-1

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

T(n)= O(n)

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