JEE Main 2023 — Mathematics Algebra

Given,

$f(x+y)=f(x)\cdot f(y)$f(x+y)=f(x)⋅f(y) and $f(1)=3$f(1)=3

Now taking x=1&y=1 we get,

$$f(1+1)=f(1)\cdot f(1)\Rightarrow f(2)={3}^{2}$$f(1+1)=f(1)⋅f(1)⇒f(2)=32

Similarly $f(2+1)=f(2)\cdot f(1)\Rightarrow f(3)={3}^{2}\times 3={3}^{3}$f(2+1)=f(2)⋅f(1)⇒f(3)=32×3=33

And so on $f(n)={3}^{n}$f(n)=3n

So, $\displaystyle \sum _{r=1}^{n}f(r)=3279$r=1∑n​f(r)=3279

$$\Rightarrow f(1)+f(2)+f(3).......f(n)=3279$$⇒f(1)+f(2)+f(3).......f(n)=3279

$$\Rightarrow 3+{3}^{2}+{3}^{3}.......+{3}^{n}=3279$$⇒3+32+33.......+3n=3279

$$\Rightarrow 3\times \frac{{3}^{n}-1}{3-1}=3279$$⇒3×3−13n−1​=3279

$$\Rightarrow \frac{{3}^{n}-1}{2}=1093$$⇒23n−1​=1093

$$\Rightarrow {3}^{n}-1=2186$$⇒3n−1=2186

$$\Rightarrow {3}^{n}=2187$$⇒3n=2187

$$\Rightarrow {3}^{n}={3}^{7}$$⇒3n=37

$$\Rightarrow n=7$$⇒n=7