Let f be a function satisfying $2f(xy)=[f(x)]^y+[f(y)]^x$ and $f(1) = K ≠1$. Then $\sum\limits_{r=1}^{n}f(r)$ is equal to
Answer & explanation
Correct answer: option 1
Since $2f(xy)=[f(x)]^y+[f(y)]^x$
Putting y = 1, we get
$2f(x)=f(x)+(f(1))^x⇒f(x)=K^x[∵f(1)=K]$
$∴\sum\limits_{r=1}^{n}f(r)=\sum\limits_{r=1}^{n}K^r=K+K^2+.....+K^n=\frac{K(K^n-1)}{K-1}$