Its GIF not GF
If function f is defined as f ( n ) = ⌊ n + 0 . 5 ⌋ , find the value of ⌊ n = 1 ∑ ∞ 2 n 2 f ( n ) + 2 − f ( n ) ⌋ .
The answer is 3.
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1 solution
- How did the strict inequality in the first two lines "loosen" out to ≤ in the third line?
- How do you get the fourth line from the third? On solving we get: k 2 − k + 0 . 2 5 ≤ n < k 2 + k + 0 . 2 5 and not what's written there.
Let [ n^1/2 + 0.5]=k
k<=n^1/2+ 0.5< k+1
(k-0.5)^2<=n<=(k+0.5)^2
on solving we get,
k^2 - k +1 <=n<=k^2+k
so
S= ∑(2^k + 2^-k)/2^n = (2+2^-1)/2 + (2^2 + 2^-2)/2^2.+...........................infinity where n= 1 to infinity
=Σ(2^-k(k-2) - 2^-k(k+2))= (2-2^-3) + (1 - 2^-8) + (2^-3 - 2^-15) +.....=3