Geometric Progression

Algebra Level 3

Find the positive integer $a$ for which $\displaystyle \sum_{k=1}^n f(a+k) = 16(2^n - 1)$ , where function $f$ satisfies $f(x+y) = f(x) \cdot f(y)$ for all positive integers $x$ and $y$ , and that $f(1) = 2$ .

The answer is 3.

1 solution

Dino Wibisono
Jan 14, 2017

We can deduce that

f(x)=2^x, that makes the equation becomes

(2^a)(Σ(2^k))=16(2^n-1), with k from 1 to n.

(2^a)(2+2^2+2^3+...+2^n)=16(2^n-1)

2(2^a)(1+2+2^2+2^3+...+2^(n-1))=16(2^n-1)

We know that

1+x+x^2+x^3+...+x^(n-1)=(x^n-1)/(x-1), so

2(2^a)(2^n-1)=16(2^n-1)

2^a=8

a=3

Geometric Progression

The answer is 3.

1 solution

0 pending reports