Question

Let $P\left( n \right):{{a}^{n}}+{{b}^{n}}$ such that a, b are even, then $P\left( n \right)$ will be divisible by $a+b$ if
A. $n>1$
B. n is odd
C. n is even
D. none of these

Answer 1

We first use the mathematical induction to find out the validation of the statement. We put successive values of n starting from 1. We try to find if $P\left( n \right)$ is divisible by $a+b$. We find the general form of n for which $P\left( n \right)$ will be divisible by $a+b$.

Complete step-by-step solution:
It’s given that $P\left( n \right):{{a}^{n}}+{{b}^{n}}$ such that a, b is even.
We use the method of mathematical induction to find out the validation of the statement.
Let’s assume $n=1$. $P\left( 1 \right)=a+b$ and that is divisible by $a+b$.
For $n=2$. $P\left( 2 \right)={{a}^{2}}+{{b}^{2}}$ and we can’t say exactly if that is divisible by $a+b$.
For $n=3$. $P\left( 3 \right)={{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$ and that is divisible by $a+b$.
For $n=4$. $P\left( 4 \right)={{a}^{4}}+{{b}^{4}}$ and we can’t say exactly if that is divisible by $a+b$.
For $n=5$. $P\left( 3 \right)={{a}^{5}}+{{b}^{5}}=\left( a+b \right)\left[ \left( {{a}^{2}}+{{b}^{2}} \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)-{{a}^{2}}{{b}^{2}} \right]$ and that is divisible by $a+b$.
So, we can see for even values of n we are not sure if $P\left( n \right)$ will be divisible by $a+b$ as we have to be exact about the values of a and b.
But for the odd values of n, we can surely say that $P\left( n \right)$ will be divisible by $a+b$.
So, we can say that $P\left( n \right):{{a}^{n}}+{{b}^{n}}$ such that a, b are even, then $P\left( n \right)$ will be divisible by $a+b$ if n is odd which means $n\in 2k+1\left[ k\in \mathbb{Z} \right]$.
The correct option is B.

Note: The well-established fact is that the sum of two odd powers whose bases are even will be always divisible by the sum of their bases. The values of a and b can change the statement and the possible option for n.