JEE Mathematics 2023 Question with Solution

Given: $5f(x+y)=f(x)\cdot f(y)$ for all $x,y\in\mathbb{R}$ and $f(3)=320$.

Find: $\sum_{n=0}^{5} f(n)$.

First, put $y=0$ in the functional equation:

$$5f(x+0)=f(x)\cdot f(0)$$

So,

$$5f(x)=f(x)\cdot f(0)$$

Since $f(x)>0$, we can divide by $f(x)$ to get

$$f(0)=5$$

Now put $y=1$:

$$5f(x+1)=f(x)\cdot f(1)$$

Hence,

$$f(x+1)=f(x)\cdot \frac{f(1)}{5}$$

Let

$$c=\frac{f(1)}{5}$$

Then

$$f(x+1)=c\,f(x)$$

Therefore, for integers $n\ge 0$,

$$f(n)=f(0)c^n=5c^n$$

Using $f(3)=320$,

$$5c^3=320$$

So,

$$c^3=64$$

which gives

$$c=4$$

Thus,

$$f(n)=5\cdot 4^n$$

Now,

$$\sum_{n=0}^{5} f(n)=5\sum_{n=0}^{5}4^n$$

Using the geometric series sum,

$$\sum_{n=0}^{5}4^n=\frac{4^6-1}{4-1}=\frac{4096-1}{3}=1365$$

Hence,

$$\sum_{n=0}^{5} f(n)=5\cdot 1365=6825$$

Therefore, the correct option is C.

Given: $5f(x+y)=f(x)f(y)$ and $f(3)=320$.

Find: $\sum_{n=0}^{5} f(n)$.

From $y=0$,

$$5f(x)=f(x)f(0)$$

so

$$f(0)=5$$

Now the relation with $y=1$ gives successive terms in integer arguments:

$$5f(x+1)=f(x)f(1)$$

Therefore,

$$f(x+1)=f(x)\cdot \frac{f(1)}{5}$$

This means $f(0),f(1),f(2),\dots$ form a geometric progression with common ratio

$$\frac{f(1)}{5}$$

Let this ratio be $r$. Then

$$f(n)=5r^n$$

Using $f(3)=320$,

$$5r^3=320$$

so

$$r^3=64=4^3$$

Hence,

$$r=4$$

Thus the required sum is

$$5(1+4+16+64+256+1024)$$

Now,

$$1+4+16+64+256+1024=1365$$

Therefore,

$$\sum_{n=0}^{5} f(n)=5\times 1365=6825$$

So the value is $6825$.