Question

Let $f$ be a function $2f\left(x,y\right)={\left[f\left(x\right)\right]}^y+{\left[f\left(y\right)\right]}^x$and $f\left(1\right)=p\ne 1$. Then $\left(p-1\right)\sum _{r=1}^{n}f\left(r\right)=$

• A. $\frac{p\left(p^n-1\right)}{\left(p-1\right)}$
• B. $\frac{p^n}{p-1}$
• C. $\frac{\left(p^n+1\right)}{\left(p+1\right)}$
• D. $p^{n+1}-p$

Answer 1

The correct option is D

$p^{n+1}-p$

Explanation for correct option:

Given data

$2f\left(x,y\right)={\left[f\left(x\right)\right]}^y+{\left[f\left(y\right)\right]}^x$

$f\left(1\right)=p\ne 1$

Substituting $y=1$, we get

$2f\left(x\right)=f\left(x\right)+{\left[f\left(1\right)\right]}^x$

From the given data $f\left(1\right)=p\ne 1$

$$\begin{gathered} 2f\left(x\right)=f\left(x\right)+{\left[f\left(1\right)\right]}^x \\ 2f\left(x\right)=f\left(x\right)+p^x \\ 2f\left(x\right)-f\left(x\right)=p^x \\ f\left(x\right)=p^x \end{gathered}$$

Then,

$$\begin{gathered} \sum _{r=1}^{n}f\left(r\right)=\sum _{r=1}^n{p}^r \\ \sum _{r=1}^{n}f\left(r\right)=\frac{p^{n+1}-p}{p-1} \\ \left(p-1\right)\sum _{r=1}^{n}f\left(r\right)=p^{n+1}-p \end{gathered}$$

Hence, the correct answer is Option (D).