f is a function for which f(1) = 1 and f(x) = 2x + f(x – 1ؘ) for each natural number x ≥ 2. Find f(31).
Explanation:
Given function can be rearranged as,
f(x) – f(x – 1) = 2x
⇒ f(2) – f(1) = 2 × 2 ⇒ f(3) – f(2) = 2 × 3 ⇒ f(4) – f(3) = 2 × 4 … ⇒ f(31) – f(30) = 2 × 31
Adding all these equations we get
f(31) – f(1) = 2 × (2 + 3 + 4 + … + 31)
f(31) – 1 = 2 × ((31×32)/2-1) = 990
⇒ f(31) = 991
Alternately, From the given function ;
f(2) = 4 + f(1) = 5 = 22 + 1
f(3) = 6 + f(2) = 6 + 5 = 11 = 32 +2
f(4) = 8 + f(3) = 8 + 11 = 19 = 42 + 3
Hence we can see that f(x) = x2 + (x-1)
Hence f(31) = 312 + 30 = 991.
Hence, option (d).
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