Practicing Success
Practicing Success
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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 |
$\frac{K(K^n-1)}{K-1}$ $\frac{K^n}{K-1}$ $\frac{K^n+1}{K+1}$ none of these |
$\frac{K(K^n-1)}{K-1}$ |
Since $2f(xy)=[f(x)]^y+[f(y)]^x$ Putting y = 1, we get $2f(x)=f(x)+(f(1))^x⇒f(x)=K^x[∵f(1)=K]$ $∴\sum\limits_{r=1}^{n}f(r)=\sum\limits_{r=1}^{n}K^r=K+K^2+.....+K^n=\frac{K(K^n-1)}{K-1}$ |
