Let _k=1^10f(a+k) = 16(2^10-1), where the function f satisfies f(x+y)=f(x)f(y) for all nat…

Question

Let $\displaystyle\sum_{k=1}^{10}f(a+k) = 16(2^{10}-1)$, where the function $f$ satisfies $f(x+y)=f(x)f(y)$ for all natural numbers $x,y$, and $f(1)=2$. Then the natural number $a$ is:

Accepted Answer

Given f(1) = 2

and f(x + y) = f(x)f(y)

When x = 1 and y = 1 then,

f(1 + 1) = f(1)f(1)

$$ \Rightarrow $$ f(2) = (f(1))2 = 22

Also when x = 2 and y = 1 then,

f(2 + 1) = f(2)f(1)

$$ \Rightarrow $$ f(3) = 23

$$ \therefore $$ Similarly f(4) = 24

.

f(x) = 2x

$$ \therefore $$ f(a + k) = 2a + k

Now given,

$$\sum\limits_{k = 1}^{10} {f(a + k) = 16\left( {{2^{10}} - 1} \right)} $$

$$ \Rightarrow $$ $$\sum\limits_{k = 1}^{10} {{2^{a + k}}} $$ = $$16\left( {{2^{10}} - 1} \right)$$

$$ \Rightarrow $$ $${2^{a + 1}} + {2^{a + 2}} + .... + {2^{a + 10}}$$ = $$16\left( {{2^{10}} - 1} \right)$$

$$ \Rightarrow $$ $${2^a}\left[ {2 + {2^2} + .... + {2^{10}}} \right]$$ =  $$16\left( {{2^{10}} - 1} \right)$$

$$ \Rightarrow $$ $${2^a} \times {{2\left( {{2^{10}} + 1} \right)} \over {2 - 1}}$$ = $$16\left( {{2^{10}} - 1} \right)$$

$$ \Rightarrow $$ ${2^a} = 8$

$$ \Rightarrow $$ $a$ = 3

Solutions · 1