JEE MAIN - Mathematics (2023 - 24th January Evening Shift - No. 1)
Let $$f(x)$$ be a function such that $$f(x+y)=f(x).f(y)$$ for all $$x,y\in \mathbb{N}$$. If $$f(1)=3$$ and $$\sum\limits_{k = 1}^n {f(k) = 3279} $$, then the value of n is
9
7
6
8
Explanation
$$
\begin{aligned}
& \mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{y}) \forall \mathrm{x}, \mathrm{y} \in \mathrm{N}, \mathrm{f}(1)=3 \\\\
& \mathrm{f}(2)=\mathrm{f}^2(1)=3^2 \\\\
& \mathrm{f}(3)=\mathrm{f}(1) \mathrm{f}(2)=3^3 \\\\
& \mathrm{f}(4)=3^4 \\\\
& \mathrm{f}(\mathrm{k})=3^{\mathrm{k}} \\\\
& \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{f}(\mathrm{k})=3279 \\\\
& \mathrm{f}(1)+\mathrm{f}(2)+\mathrm{f}(3)+\ldots \ldots \ldots+\mathrm{f}(\mathrm{k})=3279 \\\\
& 3+3^2+3^3+\ldots \ldots \ldots 3^{\mathrm{k}}=3279 \\\\
& \frac{3\left(3^{\mathrm{k}}-1\right)}{3-1}=3279 \\\\
& \frac{3^{\mathrm{k}}-1}{2}=1093 \\\\
& 3^{\mathrm{k}}-1=2186 \\\\
& 3^{\mathrm{k}}=2187 \\\\
& \text{So, } \mathrm{k}=7
\end{aligned}
$$
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