JEE MAIN - Mathematics (2020 - 6th September Evening Slot - No. 12)
Suppose that a function f : R $$ \to $$ R satisfies
f(x + y) = f(x)f(y) for all x, y $$ \in $$ R and f(1) = 3.
If $$\sum\limits_{i = 1}^n {f(i)} = 363$$ then n is equal to ________ .
f(x + y) = f(x)f(y) for all x, y $$ \in $$ R and f(1) = 3.
If $$\sum\limits_{i = 1}^n {f(i)} = 363$$ then n is equal to ________ .
Answer
5
Explanation
f(x + y) = f(x) f(y)
put x = y = 1
$$ \therefore $$ f(2) = (ƒ(1))2 = 32
put x = 2, y = 1
$$ \therefore $$ f(3) = (ƒ(1))3 = 33
Similarly f(x) = 3x
$$ \Rightarrow $$ f(i) = 3i
Given, $$\sum\limits_{i = 1}^n {f(i)} = 363$$
$$ \Rightarrow $$ 3 + 32 + 33 +.... + 3n = 363
$$ \Rightarrow $$ $${{3\left( {{3^n} - 1} \right)} \over {3 - 1}}$$ = 363
$$ \Rightarrow $$ 3n - 1 = $${{363 \times 2} \over 3}$$ = 242
$$ \Rightarrow $$ 3n = 243 = 35
$$ \Rightarrow $$ n = 5
put x = y = 1
$$ \therefore $$ f(2) = (ƒ(1))2 = 32
put x = 2, y = 1
$$ \therefore $$ f(3) = (ƒ(1))3 = 33
Similarly f(x) = 3x
$$ \Rightarrow $$ f(i) = 3i
Given, $$\sum\limits_{i = 1}^n {f(i)} = 363$$
$$ \Rightarrow $$ 3 + 32 + 33 +.... + 3n = 363
$$ \Rightarrow $$ $${{3\left( {{3^n} - 1} \right)} \over {3 - 1}}$$ = 363
$$ \Rightarrow $$ 3n - 1 = $${{363 \times 2} \over 3}$$ = 242
$$ \Rightarrow $$ 3n = 243 = 35
$$ \Rightarrow $$ n = 5
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