Courses for Kids
Free study material
Offline Centres
Store Icon

If $a,b$ and $n$ are natural numbers then ${{a}^{2n-1}}+{{b}^{2n-1}}$ is divisible by
A. $a+b$
B. $a-b$
C. ${{a}^{3}}+{{b}^{3}}$
D. ${{a}^{2}}+{{b}^{2}}$

Last updated date: 21st Jul 2024
Total views: 348.6k
Views today: 7.48k
348.6k+ views
Hint: To get the divisor of the given function that is ${{a}^{2n-1}}+{{b}^{2n-1}}$, we will find the value of the function for different values of $n$ for some numbers. Then, we will look for common factors among all of them. This common factor will be the divisor of the given function.

Complete step by step answer:
Since, $n$ is the natural numbers. We will substitute the value of $n$ in the given function with numbers. The given function is:
$\Rightarrow {{a}^{2n-1}}+{{b}^{2n-1}}$

For $n=1$, the function will be:
$\Rightarrow {{a}^{2\times 1-1}}+{{b}^{2\times 1-1}}$
Now, we will simplify it by using multiplication and then subtraction as
  & \Rightarrow {{a}^{2-1}}+{{b}^{2-1}} \\
 & \Rightarrow {{a}^{1}}+{{b}^{1}} \\
The above step can be written as:
$\Rightarrow a+b$

For $n=2$, we will get from the function:
$\Rightarrow {{a}^{2\times 2-1}}+{{b}^{2\times 2-1}}$
Now, we will be using multiplication in the above step and will get $4$ for multiplication $2$ with \[2\] as:
$\Rightarrow {{a}^{4-1}}+{{b}^{4-1}}$
And then we will subtract $1$ from $4$ and will get $3$:
$\Rightarrow {{a}^{3}}+{{b}^{3}}$
Here, we can expand it as:
\[\Rightarrow \left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)\]

Now, we will check it for $n=3$:
$\Rightarrow {{a}^{2\times 3-1}}+{{b}^{2\times 3-1}}$
After multiplication, the above step will be:
$\Rightarrow {{a}^{6-1}}+{{b}^{6-1}}$
And after subtracting $1$ from $6$, the above function will be:
$\Rightarrow {{a}^{5}}+{{b}^{5}}$
Further we will expand the above step as:
$\Rightarrow \left( a+b \right)\left( {{a}^{4}}-{{a}^{3}}b+{{a}^{2}}{{b}^{2}}-a{{b}^{3}}+{{b}^{4}} \right)$

And for substituting $n=$4, the function will be as:
$\Rightarrow {{a}^{2\times 4-1}}+{{b}^{2\times 4-1}}$
Multiplication of $2$ and $4$ will gives from the above step as:
$\Rightarrow {{a}^{8-1}}+{{b}^{8-1}}$
After subtracting $1$ from $8$, the above function will be:
$\Rightarrow {{a}^{7}}+{{b}^{7}}$
The expansion of the above step is:
$\Rightarrow \left( a+b \right)\left( {{a}^{6}}-{{a}^{5}}b+{{a}^{4}}{{b}^{2}}-{{a}^{3}}{{b}^{3}}+{{a}^{2}}{{b}^{4}}-a{{b}^{5}}+{{b}^{6}} \right)$
Similarly, we will proceed further.
As we can clearly see that there is a common factor among all the values for $n$ that is $\left( a+b \right)$.
Hence, ${{a}^{2n-1}}+{{b}^{2n-1}}$ is divisible by $\left( a+b \right)$.

So, the correct answer is “Option A”.

Note: We need to take care to expand the obtained function after applying the different value of $n$. For expansion, we can use the formula mentioned below:
$\left( {{a}^{n}}+{{b}^{n}} \right)=\left( a+b \right)\left( {{a}^{n-1}}-{{a}^{n-2}}b+{{a}^{n-3}}{{b}^{2}}-\cdot \cdot \cdot +{{a}^{2}}{{b}^{n-3}}-a{{b}^{n-2}}+{{b}^{n-1}} \right)$
Where, $a,b$ and $n$ are natural numbers.