Short Answer

Show that if $a \neq b^{d}$ and $n$ is a power of $b$ , then $f (n) = C_{1} n^{d} + C_{2} n^{\log_{b^{a}}}$ , where $C_{1 =} b^{d} c / (b^{d} - a)$ and $C_{2} = f (1) + b^{d} / (a - b^{d})$

The expression $f (n) = C_{1} n^{d} + C_{2} n^{\log_{b^{a}}}$ is proved.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Define the Induction formula

Mathematical Induction is a technique of proving a statement, theorem, or formula which is thought to be true, for each and every natural number $n$

Step 2: Proceed via induction on k

Since $n$ is a power of $b$ so, $n = b^{k}$ and $k = \log_{b} n$ for some constant $k$ .

For our base case, if $n = 1$ and $k = 0$ then: $C_{1} n^{d} + C_{2} n^{\log_{b^{a}}} = C_{1} + C_{2} = b^{d} c / (b^{d} - a) + f (1) + b^{d} c / (a - b^{d}) = f (1)$

Now, for inductive hypothesis assume true for $k$ with $n = b^{k}$ .

Next, for $n = b^{k + 1}$ :

$f (n) = a f (n / b) + c n^{d} f (n) = (b^{d} c / (b^{d} - a)) n^{d} + (f (1) + b^{d} c / (a - b^{d})) n^{\log_{b^{a}}} f (n) = C_{1} n^{d} + C_{2} n^{\log_{b^{a}}}$

And the induction is complete.

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Discrete Mathematics and its Applications

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Short Answer

Show that if $a \neq b^{d}$ and $n$ is a power of $b$ , then $f (n) = C_{1} n^{d} + C_{2} n^{\log_{b^{a}}}$ , where $C_{1 =} b^{d} c / (b^{d} - a)$ and $C_{2} = f (1) + b^{d} / (a - b^{d})$

Step by Step Solution

Step 1: Define the Induction formula

Step 2: Proceed via induction on k

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Show that if a≠bd and n is a power of b, then f(n)=C1nd+C2nlogba, where C1=bdc/(bd-a) and C2=f(1)+bd/(a-bd)

Step by Step Solution

Step 1: Define the Induction formula

Step 2: Proceed via induction on k

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Show that if $a \neq b^{d}$ and $n$ is a power of $b$ , then $f (n) = C_{1} n^{d} + C_{2} n^{\log_{b^{a}}}$ , where $C_{1 =} b^{d} c / (b^{d} - a)$ and $C_{2} = f (1) + b^{d} / (a - b^{d})$