Question

pls ans this tough question​

Answer 1

First let some examples by giving values for n be done.

Let n = 1.

$\dfrac{x^2-y^2}{x-y}=\dfrac{(x+y)(x-y)}{x-y}=x+y$

Let n = 2.

$\dfrac{x^4-y^4}{x^2-y^2}=\dfrac{(x^2+y^2)(x^2-y^2)}{x^2-y^2}=x^2+y^2$

Let n = 3.

$\dfrac{x^8-y^8}{x^4-y^4}=\dfrac{(x^4+y^4)(x^4-y^4)}{x^4-y^4}=x^4+y^4$

Now we can take n directly.

$\begin{aligned}&\dfrac{x^{2^n}-y^{2^n}}{x^{2^{n-1}}-y^{2^{n-1}}}\\ \\ \Longrightarrow\ \ &\dfrac{x^{2^{n}}-(xy)^{2^{n-1}}+(xy)^{2^{n-1}}-y^{2^{n}}}{x^{2^{n-1}}-y^{2^{n-1}}}\\ \\ \Longrightarrow\ \ &\frac{x^{2^{n-1}}(x^{2^{n-1}}-y^{2^{n-1}})+y^{2^{n-1}}(x^{2^{n-1}}-y^{2^{n-1}})}{x^{2^{n-1}}-y^{2^{n-1}}}\\ \\ \Longrightarrow\ \ &\dfrac{(x^{2^{n-1}}+y^{2^{n-1}})(x^{2^{n-1}}-y^{2^{n-1}})}{x^{2^{n-1}}-y^{2^{n-1}}}\\ \\ \Longrightarrow\ \ &x^{2^{n-1}}+y^{2^{n-1}}\end{aligned}$

Hence,

$\Large \boxed{\dfrac{x^{2^n}-y^{2^n}}{x^{2^{n-1}}-y^{2^{n-1}}}=x^{2^{n-1}}+y^{2^{n-1}}}$

If this method is quite difficult to write or understand, it will be easier to indicate 2^(n - 1) by a letter.

Let 2^(n - 1) = k.

2^n will be,

2^n = 2^(n - 1 + 1) = 2^(n - 1) × 2 = 2k.

So,

$\begin{aligned}&\dfrac{x^{2^n}-y^{2^n}}{x^{2^{n-1}}-y^{2^{n-1}}}\\ \\ \Longrightarrow\ \ &\frac{x^{2k}-y^{2k}}{x^k-y^k}\\ \\ \Longrightarrow\ \ &\frac{(x^k)^2-(y^k)^2}{x^k-y^k}\\ \\ \Longrightarrow\ \ &\frac{(x^k+y^k)(x^k-y^k)}{x^k-y^k}\\ \\ \Longrightarrow\ \ &x^k+y^k\\ \\ \Longrightarrow\ \ &x^{2^{n-1}}+y^{2^{n-1}}\end{aligned}$