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QJEEM23Ma999d: JEE Main 2023 Mathematics - Class 11 JEEM23Ma999d Solution: Let be a function such that for all . If and , then the value…

Let f(x)f(x) be a function such that f(x+y)=f(x).f(y)f(x+y)=f(x).f(y) for all x,y∈Nx,y\in \mathbb{N} . If f(1)=3f(1)=3 and ∑k=1nf(k)=3279\sum\limits_{k = 1}^n {f(k) = 3279} , then the value of n is

Multiple Choice Options

  • A 9
  • B 7
  • C 6
  • D 8

Step-by-Step Verified Solution

amp;f(x+y)=f(x)⋅f(y)∀x,y∈N,f(1)=3amp;f(2)=f2(1)=32amp;f(3)=f(1)f(2)=33amp;f(4)=34amp;f(k)=3kamp;∑k=1nf(k)=3279amp;f(1)+f(2)+f(3)+………+f(k)=3279amp;3+32+33+………3k=3279amp;3(3k−1)3−1=3279amp;3k−12=1093amp;3k−1=2186amp;3k=2187amp;So, k=7\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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