 QUESTION

# If $n\in N$, then ${{x}^{2n-1}}+{{y}^{2n-1}}$ is divisible by:A. x + yB. x - yC. ${{x}^{2}}+{{y}^{2}}$D. none of these

Hint: Take the expression equal to P(n). Put n = 1 and find the divisibility. Suppose that P(n) is true for n = k. Then prove that P(n) for n = k + 1 is true.

Let us consider $P(n)={{x}^{2n-1}}+{{y}^{2n-1}}......(1)$

First let us put n = 1 in equation (1).

\begin{align} & \therefore P(1)={{x}^{2\times 1-1}}+{{y}^{2\times 1-1}}={{x}^{2-1}}+{{y}^{2-1}}={{x}^{1}}+{{y}^{1}}=x+y \\ & \therefore P(1)=x+y. \\ \end{align}

i.e. $(x+y)$ is divisible by $(x+y)$.

Thus we can say that $P(n)$ is true for n = 1.

Suppose $P(n)$ is true for n = k.

$\therefore P(k)={{x}^{\left( 2k-1 \right)}}+{{y}^{\left( 2k-1 \right)}}$ is divisible by $(x+y)$.

Now let us check whether $P(n)$ for n = k + 1 is true or not.

Put n = k + 1 in equation (1).

\begin{align} & P\left( k+1 \right)={{x}^{2\left( k+1 \right)-1}}+{{y}^{2\left( k+1 \right)-1}} \\ & ={{x}^{2k+2-1}}+{{y}^{2k+2-1}} \\ & ={{x}^{2+\left( 2k-1 \right)}}+{{y}^{2+\left( 2k-1 \right)}} \\ \end{align}

We know the exponent product rule, when we multiply two powers that have the same base, you can add the exponents.

e.g.: ${{x}^{m}}.{{x}^{n}}={{x}^{m+n}}$

Thus we can write the expression is,

$P(k+1)={{x}^{2}}.{{x}^{2k-1}}+{{y}^{2}}.{{y}^{2k-1}}$.

Let us add and subtract ${{x}^{2}}$ in the second term.

$\therefore P(k+1)={{x}^{2}}-{{x}^{2k-1}}+\left[ {{y}^{2}}-{{x}^{2}}+{{x}^{2}} \right]{{y}^{2k-1}}$
Thus let us take ${{x}^{2}}$ common from both the terms.

\begin{align} & \therefore P(k+1)={{x}^{2}}\left[ {{x}^{2k-1}} \right]+\left[ {{y}^{2}}-{{x}^{2}} \right]{{y}^{2k-1}}+{{x}^{2}}.{{y}^{2k-1}} \\ & ={{x}^{2}}\left[ {{x}^{2k-1}}+{{y}^{2k-1}} \right]+\left( {{y}^{2}}-{{x}^{2}} \right){{y}^{2k-1}} \\ \end{align}

We know that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$

$\therefore P(k+a)={{x}^{2}}\left( {{x}^{2k-1}}+{{y}^{2k-1}} \right)+\left( y-x \right)\left( y+x \right){{y}^{2k-1}}$

Thus we can say that P(k + 1) is divisible by (x + y).

We proved $P(k)={{x}^{2k-1}}+{{y}^{2k-1}}$ is divisible by (x + y).

$\therefore P(n)$ for n = k + 1 is true.

Hence we can say that ${{x}^{\left( 2n-1 \right)}}+{{y}^{\left( 2n-1 \right)}}$ is divisible by (x + y), $n\in N$.

Option A is the correct answer.

Note: If you put n = 0 instead of n = 1, then you get $P(n)={}^{1}/{}_{x}+{}^{1}/{}_{y}$. But we don’t have this option in our circle and it is wrong. To prove divisibility in any question similar to it we must take, n = 1, n = k and n = k + 1. This is the standard we have to follow for similar questions.