Let ∑k=1 to10 f(a+k)=16(2¹⁰−1)... | Functions | JEE Main

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JEE MAIN 2019

09-04-2019 - S1

Question

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

Select the correct option:

A

2

B

3

C

16

D

4

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

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