Sequences and Series · Question 1

Let R denote the set of all real numbers. Let f: R R be a function such that f(x) > 0 for all x R, and f(x + y) = f(x)f(y) for all x, y R. Let the real numbers a1, a2, , a50 be in an arithmetic progression. If f(a31) = 64f(a25), and i=150 f(ai) = 3(225 + 1), then the value of i=630 f(ai) is .

Answer

96

Solution

With common difference d: f(ai+1) = f(ai + d) = f(ai)f(d), so f(a1), f(a2), is a G.P. with ratio r = f(d) > 0.

f(a31) = r6f(a25) = 64f(a25) r = 2.

i=150 f(ai) = A(250 - 1) = A(225 - 1)(225 + 1) = 3(225 + 1) A = 3225 - 1, where A = f(a1).

i=630 f(ai) = A 25(1 + 2 + + 224) = A 25(225 - 1) = 3 32 = 96.

Answer: 96