Solve the recurrence relation using iterative method subject to the basis step [13 points] s(1)=1 s(n)=s(n-1)+(2n-1),for n≥2 Then, verify the solution by using mathematical induction [7 points]
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Given s(n)=s(n-1)+(2n-1)
So,
s(n)=(s(n-2)+(2n-3))+2n-1
So,
s(n)=s(n-3)+(2n-5)+(2n-3)+(2n-1)
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s(n)=s(n-(n-1))+(2n)*(n-1)-(1+3+5....n-1 terms)
So,
s(n)=1+2n*(n-1)-((n-1)/2)*(2+(n-2)*2)=n^2
So,
s(n)=n^2
Base case: n=1
So,
s(1)=1
which is trrue
Now consider for kth term as true
s(k)=k^2
So, for
s(k+1)=s(k)+2k+1=k^2+2k+1=(k+1)^2
Which is true
So, we can say that it is true for all n
Proof by induction
Kindly revert for any queries
Thanks.
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