Question

If \[n\in N\], then \[{{x}^{2n-1}}+{{y}^{2n-1}}\] is divisible by:
A. x + y
B. x - y
C. \[{{x}^{2}}+{{y}^{2}}\]
D. none of these

Answer 1

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.

Complete step-by-step answer:
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.