Let f : R → R be a function which satisfies
f(x + y) = f(x) + f(y) ∀ x, y ∈ R. If f(1) = 2 and
g(n) = Σₖ₌₁⁽ⁿ⁻¹⁾ f( k ), n ∈ N then the value of n, for which g(n) = 20, is :

Solution

Maths sir · 4 steps

Find f at whole numbers first: adding 1 again and again gives f(k) = 2k, then the sum is easy.

1. Values at integers. Put y = 1: f(x+1) = f(x) + f(1) = f(x) + 2. Starting from f(1) = 2, each step up by 1 adds 2, so f(k) = 2k for every natural number k.
2. Sum the series. The sum stops at k = n-1, not at n. Use 1 + 2 + … + m = m(m+1)/2 with m = n-1.

   g(n) = Σₖ₌₁ⁿ⁻¹ 2k = 2 × (n-1)n/2 = n(n-1)

3. Solve. n(n-1) = 20 gives n² - n - 20 = 0, that is (n-5)(n+4) = 0. The roots are 5 and -4, and n must be a natural number, so n = -4 is rejected.
4. So. n = 5. Check: g(5) = f(1)+f(2)+f(3)+f(4) = 2+4+6+8 = 20. The answer is (c).

The idea to keep. For f(x+y) = f(x) + f(y), putting y = 1 again and again gives f(k) = k f(1) at whole numbers. After that it is only a series sum, so read the upper limit carefully.

Answer from the NTA final answer key. Solution by PrepWiser.