Question

Prove that $x\left( {{x}^{n-1}}-n{{a}^{n-1}} \right)+{{a}^{n}}\left( n-1 \right)$ is divisible by ${{\left( x-a \right)}^{2}}$ for all positive integers n greater than 1

Answer 1

Hint: We will be using mathematical induction here. In mathematical induction we substitute some value at the beginning, if that value satisfies our equation then we will substitute some variable say ‘k’ which will be in our range. Now, we assume that our function is satisfied when we substitute ‘k’ and then we prove that our function satisfies when substitute ‘k+1’. Suppose our function is $f\left( x \right)$where $x>1$, so we start by substituting the first value as ‘2’. Now if ‘2’ satisfies our function $f\left( x \right)$then we assume that our function is satisfied by a variable say ‘k’ and then prove that our function is satisfied when we substitute ‘k+1’. For a function to be completely divisible by some factor, that factor should be a part of the factors of the function.

Complete step-by-step answer:
We will start by substituting the value of as 2.
So, the equation will become,
$\begin{align}
  & f\left( n \right)=x\left( {{x}^{n-1}}-n{{a}^{n-1}} \right)+{{a}^{n}}\left( n-1 \right) \\
 & f\left( 2 \right)=x\left( {{x}^{2-1}}-2{{a}^{2-1}} \right)+{{a}^{2}}\left( 2-1 \right) \\
 & =x\left( {{x}^{1}}-2{{a}^{1}} \right)+{{a}^{2}}\left( 1 \right) \\
 & =x\left( x-2a \right)+{{a}^{2}} \\
 & ={{x}^{2}}-2ax+{{a}^{2}} \\
 & ={{\left( x-a \right)}^{2}} \\
\end{align}$
i.e. $x\left( {{x}^{n-1}}-n{{a}^{n-1}} \right)+{{a}^{n}}\left( n-1 \right)$ is divisible by ${{\left( x-a \right)}^{2}}$ when n = 2.
Now, we will assume $f\left( n \right)$ is divisible by ${{\left( x-a \right)}^{2}}$ for $k>2$
So, assume,
$\begin{align}
  & f\left( k \right)=x\left( {{x}^{k-1}}-n{{a}^{k-1}} \right)+{{a}^{k}}\left( k-1 \right) \\
 & ={{\left( x-a \right)}^{2}}g\left( x \right) \\
\end{align}$
That is $x\left( {{x}^{k-1}}-n{{a}^{k-1}} \right)+{{a}^{k}}\left( k-1 \right)$$={{\left( x-a \right)}^{2}}g\left( x \right)$
Now let’s prove that$f\left( k+1 \right)$ is divisible by ${{\left( x-a \right)}^{2}}$ for $k>2$.
Let’s say
$\begin{align}
  & f\left( k+1 \right)=x\left( {{x}^{k}}-\left( k+1 \right){{a}^{k}} \right)+{{a}^{k+1}}\left( k \right) \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=x\times {{x}^{k}}-\left( k+1 \right)x{{a}^{k}}+m{{a}^{k+1}} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=x\left[ kx{{a}^{k-1}}-{{a}^{k}}\left( k-1 \right)+{{\left( x+a \right)}^{2}}g\left( x \right) \right]-\left( k+1 \right)x{{a}^{k}}+k{{a}^{k+1}} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=k{{a}^{k-1}}\left[ \left( {{x}^{2}}-2xa+{{a}^{2}} \right)+x{{\left( x-a \right)}^{2}}g\left( x \right) \right] \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=k{{a}^{k-1}}\left( {{\left( x-a \right)}^{2}}+x{{\left( x-a \right)}^{2}}g\left( x \right) \right) \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,={{\left( x-a \right)}^{2}}\left[ k{{a}^{k-1}}+xg\left( x \right) \right] \\
\end{align}$
Hence, $f\left( k+1 \right)$ is divisible by ${{\left( x-a \right)}^{2}}$.
Therefore, by mathematical induction, $f\left( n \right)$ is divisible by ${{\left( x-a \right)}^{2}}$ for all positive integers n greater than 1.

Note: While using mathematical induction, we assume that the function is divisible by some factor, After that when we prove that the function is divisible by some value; this value we consider by addition or subtraction of 1 from the previous assumed value. Also, to check whether a function is divisible by some factor or not, use the long division of polynomials.