Question

Let ¹⁰∑ₖ₌₁ f(a + k) = 16(2¹⁰ - 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:
(A) 4 (B) 16
(C) 2 (D) 3

Answer 1

Option D: 3 is the natural number

Explanation:

Given that $\sum_{\mathrm{k}=1}^{10} \mathrm{f}(\mathrm{a}+\mathrm{k})=16\left(2^{10}-1\right)$ where the function f satisfies $f(x+y)=f(x) f(y)$ for all natural numbers x,y and $f(1)=2$

We need to determine the natural number ‘a’

When $x=1$, $y=1$, then $f(x+y)=f(x) f(y)$ becomes

$f(2)=f(1) \cdot f(1)$

$f(2)=2^2$

Similarly, when $x=2, y=1$, we have,

$f(3)=f(2) \cdot f(1)=2^{3}$

Thus, the value of x and y goes on and the general term is $f(n)=2^{n}$ ----(1)

Let us consider $\sum_{\mathrm{k}=1}^{10} \mathrm{f}(\mathrm{a}+\mathrm{k})=16\left(2^{10}-1\right)$

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

Using equation (1), we get,

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

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

Using the sum of terms, we get,

$2^{\mathrm{a}}\left[\frac{2\left(2^{10}-1\right)}{2-1}\right]=16\left(2^{10}-1\right)$

Simplifying, we get,

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

             $2^{a+1}=16$

             $2^{a+1}=2^{3+1}$

Thus, the value of the natural number a is 3

Hence, Option D is the correct answer.

Learn more:

(1) The sum of all natural numbers a such that a 2 16a + 67 is a perfect square is what?

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(2) Find the value of a+b+c+d if the product of first 10 natural numbers is written as 2a+3b+5c+7d

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