Question

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]

Answer 1

`Hey,

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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

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