Question

Show that if n is an integer, then\(\sum\limits_{{\rm{k = 0}}}^{\rm{n}} {{{\rm{3}}^{\rm{k}}}\left( {\frac{{\rm{n}}}{{\rm{k}}}} \right)} {\rm{ = 4n}}\).

Short answer

It is shown that\(\sum\limits_{k = 0}^n {{3^k}\left( {\frac{n}{k}} \right)} = 4n\).

Step-by-step solution

Introduction

An integer is a number from the set of negative and positive numbers, including zero that has no decimal or fractional element.

Explanation

Suppose n is an integer then,

\(\begin{array}{c}\sum\limits_{{\rm{k = 0}}}^{\rm{n}} {{{\rm{3}}^{\rm{k}}}} {\rm{ \times }}\left( {\begin{array}{*{20}{l}}{\rm{n}}\\{\rm{k}}\end{array}} \right)\\{\rm{ = }}\sum\limits_{{\rm{k = 0}}}^{\rm{n}} {\left( {\begin{array}{*{20}{l}}{\rm{n}}\\{\rm{k}}\end{array}} \right)} {\rm{ \times }}{{\rm{1}}^{{\rm{n - k}}}}{\rm{ \times }}{{\rm{3}}^{\rm{k}}}\\{\rm{ = (1 + 3}}{{\rm{)}}^{\rm{n}}}\\{\rm{ = }}{{\rm{4}}^{\rm{n}}}\end{array}\)

Hence, it is shown that\(\sum\limits_{k = 0}^n {{3^k}\left( {\frac{n}{k}} \right)} = 4n\).